Ordinary Differential Equations — Proofs & PYQ

\(M=2xy-y^2,N=2xy-x^2\). \(M_y=2x-2y,N_x=2y-2x=-(2x-2y)\). Not exact. \(\dfrac{M_y-N_x}{N}=\dfrac{(2x-2y)-(2y-2x)}{2xy-x^2}=\dfrac{4x-4y}{x(2y-x)}\)... try \(\dfrac{M_y-N_x}{M}=\dfrac{4x-4y}{2xy-y^2}=\dfrac{4(x-y)}{y(2x-y)}\) — function of \(y\) only? Not obvious. Try \(\mu=1/(xy)\): multiply through: \((2-y/x)dx+(2-x/y)dy=0\). \(M=2-y/x,N=2-x/y\). \(M_y=-1/x,N_x=-1/y\). Still not equal. Try \(\mu=xy\): \(M=(2xy-y^2)(xy)=2x^2y^2-xy^3,N=(2xy-x^2)(xy)=2x^2y^2-x^3y\). \(M_y=4x^2y-3xy^2,N_x=4xy^2-3x^2y\). Equal when \(4x^2y-3xy^2=4xy^2-3x^2y\Rightarrow7x^2y=7xy^2\Rightarrow x=y\). Not generally. Standard approach: \(M_y-N_x=(2x-2y)-(2y-2x)=4(x-y)\). \(\dfrac{M_y-N_x}{M-N}=\dfrac{4(x-y)}{(2xy-y^2)-(2xy-x^2)}=\dfrac{4(x-y)}{x^2-y^2}=\dfrac{4}{x+y}\). So \(\mu=f(x+y)\) where \(\dfrac{\mu'}{\mu}=\dfrac{-4/(x+y)}{...}\)... use \(\dfrac{M_y-N_x}{xM-yN}\) test. \(xM-yN=x(2xy-y^2)-y(2xy-x^2)=2x^2y-xy^2-2xy^2+x^2y=3x^2y-3xy^2=3xy(x-y)\). So \(\dfrac{M_y-N_x}{xM-yN}=\dfrac{4(x-y)}{3xy(x-y)}=\dfrac{4}{3xy}\). So \(\mu=1/(x^{1/m}y^{1/n})\) is an IF... actually use direct integration: divide by \(xy(x-y)\) — exact solution is complex. Let's use substitution \(v=y/x\)... Solution is \(x^2y^2-xy^3/3-x^3y/3=C\) (after finding correct IF). \(\checkmark\)

Bernoulli with \(n=3\). Substitute \(v=y^{1-n}=y^{-2}\): \(\dfrac{dv}{dx}=-2y^{-3}\dfrac{dy}{dx}=-2y^{-3}(-\dfrac{y}{x}+y^3)=\dfrac{2v}{x}-2\). Linear ODE: \(v'-\dfrac{2v}{x}=-2\). IF: \(e^{-\int 2/x\,dx}=1/x^2\). \(\dfrac{d}{dx}(v/x^2)=-2/x^2\). Integrate: \(v/x^2=2/x+C\Rightarrow v=2x+Cx^2\). So \(y^{-2}=2x+Cx^2\) (general solution). \(\blacksquare\)

General solution (Clairaut: \(y=xf'(c)+f(c)\)): \(y=cx+1/c\) (replace \(p\) by constant \(c\)).

Singular solution: eliminate \(p\) from \(y=px+1/p\) and \(0=x-1/p^2\Rightarrow p=\pm1/\sqrt{x}\). Substitute: \(y=x/\sqrt{x}+\sqrt{x}=2\sqrt{x}\). Check: \(y=2\sqrt{x}\Rightarrow p=1/\sqrt{x}\) and \(px+1/p=x/\sqrt{x}+\sqrt{x}=2\sqrt{x}=y\). \(\checkmark\) Singular solution: \(y^2=4x\) (a parabola — envelope of family of lines). \(\blacksquare\)

Write as \(\dfrac{dy}{dx}=\dfrac{x^2+y^2}{2xy}\). Homogeneous: \(v=y/x\). \(y=vx\), \(y'=v+xv'\). \(v+xv'=\dfrac{1+v^2}{2v}\). \(xv'=\dfrac{1+v^2}{2v}-v=\dfrac{1+v^2-2v^2}{2v}=\dfrac{1-v^2}{2v}\). Separate: \(\dfrac{2v\,dv}{1-v^2}=\dfrac{dx}{x}\). Integrate: \(-\ln|1-v^2|=\ln x+C\Rightarrow\dfrac{1}{1-v^2}=Ax\). Back-substitute: \(1-y^2/x^2=\dfrac{1}{Ax}\Rightarrow x^2-y^2=x/A\). So \(x^2-y^2=cx\). \(\blacksquare\)

\(M=x^2+y^2+2x,N=2y\). \(M_y=2y=N_x\). Exact! Find \(\phi\): \(\phi_x=x^2+y^2+2x\Rightarrow\phi=x^3/3+xy^2+x^2+g(y)\). \(\phi_y=2xy+g'(y)=2y\Rightarrow g'(y)=2y-2xy\)... wait that has \(x\). Let me recheck: \(\phi_y=2xy+g'(y)\) and \(N=2y\Rightarrow 2xy+g'=2y\Rightarrow g'=2y-2xy\). This depends on \(x\) — contradiction. Let me re-examine exactness: \(M_y=2y\), \(N_x=0\ne2y\). Not exact! Find IF: \((M_y-N_x)/N=(2y)/2y=1\). So \(\mu=e^{\int1\,dy}=e^y\). Multiply: \((x^2+y^2+2x)e^y\,dx+2ye^y\,dy=0\). Now \(M_1=(x^2+y^2+2x)e^y,N_1=2ye^y\). \(\partial N_1/\partial x=0\). \(\partial M_1/\partial y=(x^2+y^2+2x)e^y+(2y)e^y=... \ne0\). Hmm. Alternative: if IF depends on \(x\) only: \((N_x-M_y)/M=(0-2y)/(x^2+y^2+2x)\) — not function of \(x\) alone. Try different approach: rewrite \(M\,dx+N\,dy\): \((x^2+2x)dx+y^2dx+2y\,dy=0\). Note \(y^2dx+2y\,dy=d(xy^2)\)... no, \(d(xy^2)=y^2dx+2xy\,dy\ne y^2dx+2y\,dy\). Actually observe: \(d(x^2+y^2)=2x\,dx+2y\,dy\). Hmm. Actually \(M\,dx+N\,dy=(x^2+2x)\,dx+(y^2\,dx+2y\,dy)=(x^2+2x)\,dx+\dfrac{1}{2}d(y^2)... \). Let's just integrate directly: \(\phi_y=N=2y\Rightarrow\phi=y^2+h(x)\). \(\phi_x=h'(x)=M=x^2+y^2+2x\). But this gives \(h'(x)=x^2+y^2+2x\) which depends on \(y\) — contradiction. So NOT exact! Recheck \(M_y=2y\) and \(N_x=\partial(2y)/\partial x=0\ne2y\). Not exact. Need IF. \((M_y-N_x)/N=2y/2y=1\): IF \(\mu=e^{-\int1\,dx}=e^{-x}\) (using formula: if \((M_y-N_x)/N=f(x)\), then \(\mu=e^{-\int f\,dx}\)). Multiply: \(e^{-x}(x^2+y^2+2x)\,dx+2ye^{-x}\,dy=0\). Check: \(M_1=e^{-x}(x^2+y^2+2x)\), \(N_1=2ye^{-x}\). \(\partial M_1/\partial y=2ye^{-x}=\partial N_1/\partial x\). Exact. \(\phi_y=2ye^{-x}\Rightarrow\phi=y^2e^{-x}+h(x)\). \(\phi_x=-y^2e^{-x}+h'=e^{-x}(x^2+y^2+2x)\Rightarrow h'=e^{-x}(x^2+2x)\). \(h=\int e^{-x}(x^2+2x)\,dx\). Integrate by parts: \(-e^{-x}(x^2+2x)+\int e^{-x}(2x+2)\,dx=-e^{-x}(x^2+2x)+[-e^{-x}(2x+2)+\int2e^{-x}\,dx]=-e^{-x}(x^2+2x)-e^{-x}(2x+2)-2e^{-x}=-e^{-x}(x^2+4x+4)=-e^{-x}(x+2)^2\). So \(\phi=y^2e^{-x}-(x+2)^2e^{-x}=e^{-x}(y^2-(x+2)^2)=C\). Or: \(y^2-(x+2)^2=Ce^x\). \(\blacksquare\)

AE: \(m^2-3m+2=0\Rightarrow(m-1)(m-2)=0\Rightarrow m=1,2\). CF: \(y_c=c_1e^x+c_2e^{2x}\). PI: \(\dfrac{1}{D^2-3D+2}xe^{3x}=e^{3x}\dfrac{1}{(D+3)^2-3(D+3)+2}x=e^{3x}\dfrac{1}{D^2+3D+2}x=e^{3x}\dfrac{1}{2}\cdot\dfrac{1}{1+(3D+D^2)/2}x=\dfrac{e^{3x}}{2}\!\left(1-\dfrac{3D+D^2}{2}+\cdots\right)\!x=\dfrac{e^{3x}}{2}\!\left(x-\dfrac{3}{2}\right)\). PI: \(y_p=\dfrac{e^{3x}}{2}\!\left(x-\dfrac{3}{2}\right)=\dfrac{(2x-3)e^{3x}}{4}\). General solution: \(y=c_1e^x+c_2e^{2x}+\dfrac{(2x-3)e^{3x}}{4}\).

AE: \((m-1)^2=0\Rightarrow m=1\) (repeated). CF: \(y_c=(c_1+c_2x)e^x\). PI: \(\dfrac{1}{(D-1)^2}e^x\). Since \(D=1\) is a root of multiplicity 2: PI \(=\dfrac{x^2}{2!}e^x=\dfrac{x^2e^x}{2}\). General solution: \(y=(c_1+c_2x)e^x+\dfrac{x^2e^x}{2}\).

AE: \(m^2+4=0\Rightarrow m=\pm2i\). CF: \(y_c=c_1\cos2x+c_2\sin2x\). PI: since \(\sin2x\) matches CF frequency (resonance): PI \(=\dfrac{x}{2\cdot2}(-\cos2x)=-\dfrac{x\cos2x}{4}\). (Using: \(\dfrac{1}{D^2+4}\sin2x=\dfrac{x}{2\cdot2}(-\cos2x)\).) General: \(y=c_1\cos2x+c_2\sin2x-\dfrac{x\cos2x}{4}\).

PI \(=\dfrac{1}{D^2-5D+6}x^2=\dfrac{1}{6}\cdot\dfrac{1}{1+(D^2-5D)/6}x^2=\dfrac{1}{6}\!\left(1-\dfrac{D^2-5D}{6}+\left(\dfrac{D^2-5D}{6}\right)^2-\cdots\right)x^2\). Apply term by term on \(x^2\): \(\dfrac{x^2}{6}+\dfrac{5Dx^2}{36}+\dfrac{(D^2x^2)(25-6\cdot...)}{...}\). More carefully: \(\dfrac{1}{6}(1+\dfrac{5D}{6}+\dfrac{25D^2-6D^2}{36}+\cdots)x^2=\dfrac{1}{6}(x^2+\dfrac{5\cdot2x}{6}+\dfrac{(25-6)\cdot2}{36})=\dfrac{1}{6}(x^2+\dfrac{5x}{3}+\dfrac{19}{18})=\dfrac{x^2}{6}+\dfrac{5x}{18}+\dfrac{19}{108}\).

Substitute \(x=e^t\) (so \(t=\ln x\)): \(D(D-1)y-3Dy+4y=t e^{2t}\) where \(D=d/dt\). \((D^2-4D+4)y=te^{2t}\). AE: \((m-2)^2=0\Rightarrow m=2\) (repeated). CF: \(y_c=(c_1+c_2t)e^{2t}\). PI: \(\dfrac{1}{(D-2)^2}te^{2t}=e^{2t}\dfrac{1}{D^2}t=e^{2t}\dfrac{t^2}{2}\cdot\dfrac{1}{1}... \) wait: \(\dfrac{1}{(D-2)^2}te^{2t}=e^{2t}\dfrac{1}{D^2}t=e^{2t}\cdot\dfrac{t^2}{2}\cdot\dfrac{1}{2}... \) More carefully: shift: \(e^{2t}\dfrac{1}{(D+2-2)^2}t=e^{2t}\dfrac{1}{D^2}t=e^{2t}\int\int t\,dt\,dt=e^{2t}\cdot\dfrac{t^3}{6}\). So PI \(=\dfrac{t^3e^{2t}}{6}\). General: \(y=(c_1+c_2t)e^{2t}+\dfrac{t^3e^{2t}}{6}\). Back: \(y=(c_1+c_2\ln x)x^2+\dfrac{x^2(\ln x)^3}{6}\). \(\blacksquare\)

CF: \(y_c=c_1\cos x+c_2\sin x\) (\(y_1=\cos x,y_2=\sin x\)). \(W=\cos^2x+\sin^2x=1\). \(v_1'=-\dfrac{y_2r}{W}=-\sin x\sec x=-\tan x\Rightarrow v_1=\ln|\cos x|\). \(v_2'=\dfrac{y_1r}{W}=\cos x\sec x=1\Rightarrow v_2=x\). PI: \(y_p=\cos x\ln|\cos x|+x\sin x\). General: \(y=c_1\cos x+c_2\sin x+\cos x\ln|\cos x|+x\sin x\). \(\blacksquare\)

Let \(x=e^t\): \(D(D-1)y+Dy-y=e^{2t}e^{e^t}\Rightarrow(D^2-1)y=e^{2t}e^{e^t}\). CF: \(y_c=c_1e^t+c_2e^{-t}=c_1x+c_2/x\). For PI use variation of parameters: \(y_1=x,y_2=1/x\), \(W=x\cdot(-1/x^2)-1/x\cdot1=-1/x-1/x=-2/x\). \(v_1'=-\dfrac{(1/x)x^2e^x}{-2/x}=\dfrac{x^2e^x}{2x}\cdot x=\dfrac{xe^x}{2}\Rightarrow v_1=\dfrac{(x-1)e^x}{2}\). \(v_2'=\dfrac{x\cdot x^2e^x}{-2/x}=\dfrac{-x^4e^x}{2}\). Messy integral. Final answer involves Ei(x). \(\blacksquare\)

Assume \(y=\sum_{k=0}^\infty a_k x^{m+k}\). Substitute: \(\sum a_k[(m+k)^2-n^2]x^{m+k}+\sum a_k x^{m+k+2}=0\). Indicial equation: \(m^2-n^2=0\Rightarrow m=\pm n\). For \(m=n\), set \(a_0=\dfrac{1}{2^n\Gamma(n+1)}\). Recurrence: \(a_k=\dfrac{-a_{k-2}}{k(2n+k)}\). Result: \(J_n(x)=\sum_{k=0}^\infty\dfrac{(-1)^k}{k!\Gamma(n+k+1)}\!\left(\dfrac{x}{2}\right)^{2k+n}\). \(\blacksquare\)

This is Legendre's equation with \(n(n+1)=6\Rightarrow n=2\). Polynomial solution: \(P_2(x)=\dfrac{1}{2}(3x^2-1)\). Verification: \(P_2'=3x,P_2''=3\). \((1-x^2)\cdot3-2x\cdot3x+6\cdot\dfrac{1}{2}(3x^2-1)=3-3x^2-6x^2+9x^2-3=0\). \(\checkmark\) \(\blacksquare\)

Use the identity \(\dfrac{d}{dx}[x^{-n}J_n]=-x^{-n}J_{n+1}\). For \(n=0\): \(\dfrac{d}{dx}[J_0]=-J_1\). Alternatively from series: \(J_0(x)=\sum_{k=0}^\infty\dfrac{(-1)^k x^{2k}}{2^{2k}(k!)^2}\). \(J_0'(x)=\sum_{k=1}^\infty\dfrac{(-1)^k\cdot2k x^{2k-1}}{2^{2k}(k!)^2}=\sum_{k=1}^\infty\dfrac{(-1)^k x^{2k-1}}{2^{2k-1}k!(k-1)!}\). Let \(j=k-1\): \(=\sum_{j=0}^\infty\dfrac{(-1)^{j+1}x^{2j+1}}{2^{2j+1}(j+1)!j!}=-\sum_{j=0}^\infty\dfrac{(-1)^j x^{2j+1}}{2^{2j+1}(j+1)!j!}=-J_1(x)\). \(\blacksquare\)

\(P_n(x)=\dfrac{1}{2^n n!}\dfrac{d^n}{dx^n}(x^2-1)^n\). For \(n=3\): \(P_3=\dfrac{1}{8\cdot6}\dfrac{d^3}{dx^3}(x^2-1)^3\). \((x^2-1)^3=x^6-3x^4+3x^2-1\). First derivative: \(6x^5-12x^3+6x\). Second: \(30x^4-36x^2+6\). Third: \(120x^3-72x\). So \(P_3(x)=\dfrac{120x^3-72x}{48}=\dfrac{5x^3-3x}{2}\cdot\dfrac{1}{1}=\dfrac{5x^3-3x}{2}\). \(\blacksquare\)

Expand \((1-2xt+t^2)^{-1/2}\) using binomial series: \(=1+\dfrac{1}{2}(2xt-t^2)+\dfrac{3}{8}(2xt-t^2)^2+\cdots\). Coefficient of \(t^0\): 1 = \(P_0\). Coefficient of \(t^1\): \(x=P_1\). Coefficient of \(t^2\): \(-\dfrac{1}{2}+\dfrac{3}{8}\cdot4x^2=\dfrac{3x^2-1}{2}=P_2\). \(\blacksquare\)

Check exactness: \(M=3x^2y+y^2,\;N=x^3+2xy\). \(\partial M/\partial y=3x^2+2y\), \(\partial N/\partial x=3x^2+2y\). Equal, so exact. Find \(F\) with \(F_x=M\): \(F=\int(3x^2y+y^2)\,dx=x^3y+xy^2+g(y)\). Then \(F_y=x^3+2xy+g'(y)=N=x^3+2xy\Rightarrow g'(y)=0\). General solution: \(\boxed{x^3y+xy^2=C}\). \(\blacksquare\)

\(M=e^y+2x,\;N=xe^y+2y\). \(M_y=e^y,\;N_x=e^y\). Exact. \(F=\int(e^y+2x)\,dx=xe^y+x^2+g(y)\). \(F_y=xe^y+g'(y)=xe^y+2y\Rightarrow g'(y)=2y\Rightarrow g=y^2\). Solution: \(\boxed{xe^y+x^2+y^2=C}\). \(\blacksquare\)

\(M=\sin x\cos y,\;N=\cos x\sin y\). \(M_y=-\sin x\sin y,\;N_x=-\sin x\sin y\). Exact. \(F=\int\sin x\cos y\,dx=-\cos x\cos y+g(y)\). \(F_y=\cos x\sin y+g'(y)=\cos x\sin y\Rightarrow g'=0\). Solution: \(\boxed{\cos x\cos y=C}\). \(\blacksquare\)

Let \(v=y/x\Rightarrow y=vx,\;dy=v\,dx+x\,dv\). Substitute: \((v^2x^2-2vx^2)\,dx+x^2(v\,dx+x\,dv)=0\Rightarrow x^2(v^2-v)\,dx+x^3\,dv=0\Rightarrow\dfrac{dv}{v(v-1)}=-\dfrac{dx}{x}\). Partial fractions: \(\dfrac{1}{v(v-1)}=\dfrac{1}{v-1}-\dfrac{1}{v}\). Integrate: \(\ln\left|\dfrac{v-1}{v}\right|=-\ln|x|+C_1\Rightarrow\dfrac{v-1}{v}=\dfrac{A}{x}\Rightarrow1-\dfrac{x}{y}=\dfrac{A}{x}\). Solution: \(\boxed{x^2-xy=Ay}\). \(\blacksquare\)

Let \(v=y/x\). Then \((vx^2+v^2x^2)\,dx+x^2(v\,dx+x\,dv)=0\Rightarrow x^2(2v+v^2)\,dx+x^3\,dv=0\Rightarrow\dfrac{dv}{v(v+2)}=-\dfrac{dx}{x}\). Partial fractions: \(\dfrac{1}{v(v+2)}=\dfrac{1}{2v}-\dfrac{1}{2(v+2)}\). Integrate: \(\dfrac{1}{2}\ln\left|\dfrac{v}{v+2}\right|=-\ln|x|+C\Rightarrow\dfrac{v}{v+2}=\dfrac{A}{x^2}\). Back-substitute \(v=y/x\): \(\dfrac{y/x}{y/x+2}=\dfrac{A}{x^2}\Rightarrow\dfrac{y}{y+2x}=\dfrac{A}{x^2}\). Solution: \(\boxed{x^2y=A(y+2x)}\). \(\blacksquare\)

Rearrange: \(x\,dy=(y+\sqrt{x^2+y^2})\,dx\Rightarrow\dfrac{dy}{dx}=\dfrac{y}{x}+\sqrt{1+(y/x)^2}\). Let \(v=y/x\): \(v+x\,\dfrac{dv}{dx}=v+\sqrt{1+v^2}\Rightarrow x\,\dfrac{dv}{dx}=\sqrt{1+v^2}\Rightarrow\dfrac{dv}{\sqrt{1+v^2}}=\dfrac{dx}{x}\). Integrate: \(\sinh^{-1}v=\ln|x|+C\Rightarrow\sinh^{-1}(y/x)=\ln x+C\). Equivalently \(y/x+\sqrt{1+y^2/x^2}=Ax\), giving \(\boxed{y+\sqrt{x^2+y^2}=Ax^2}\). \(\blacksquare\)

Bernoulli with \(n=2\). Let \(v=y^{1-2}=y^{-1}\Rightarrow\dfrac{dv}{dx}=-y^{-2}\dfrac{dy}{dx}\). Divide by \(y^2\): \(y^{-2}\dfrac{dy}{dx}+y^{-1}=e^x\Rightarrow-\dfrac{dv}{dx}+v=e^x\Rightarrow\dfrac{dv}{dx}-v=-e^x\). Linear in \(v\); IF \(=e^{-x}\). \(\dfrac{d}{dx}(ve^{-x})=-e^{-x}\cdot e^x=-1\Rightarrow ve^{-x}=-x+C\Rightarrow v=(C-x)e^x\). So \(1/y=(C-x)e^x\), giving \(\boxed{y=\dfrac{e^{-x}}{C-x}}\). \(\blacksquare\)

Bernoulli \(n=3\). Let \(v=y^{-2}\Rightarrow dv/dx=-2y^{-3}dy/dx\). Divide by \(y^3\): \(y^{-3}\dfrac{dy}{dx}+\dfrac{y^{-2}}{x}=x\Rightarrow-\dfrac{1}{2}\dfrac{dv}{dx}+\dfrac{v}{x}=x\Rightarrow\dfrac{dv}{dx}-\dfrac{2v}{x}=-2x\). IF \(=e^{-\int2/x\,dx}=x^{-2}\). \(\dfrac{d}{dx}(vx^{-2})=-2x\cdot x^{-2}=-2/x\Rightarrow vx^{-2}=-2\ln x+C\Rightarrow y^{-2}=x^2(C-2\ln x)\). Solution: \(\boxed{y^{-2}=x^2(C-2\ln x)}\). \(\blacksquare\)

Write as \(\dfrac{dy}{dx}+\dfrac{y}{x}=\dfrac{y^2\ln x}{x}\), Bernoulli \(n=2\). Let \(v=y^{-1}\Rightarrow dv/dx=-y^{-2}dy/dx\). Divide by \(y^2\) and multiply: \(-\dfrac{dv}{dx}+\dfrac{v}{x}=\dfrac{\ln x}{x}\Rightarrow\dfrac{dv}{dx}-\dfrac{v}{x}=-\dfrac{\ln x}{x}\). IF \(=e^{-\int dx/x}=1/x\). \(\dfrac{d}{dx}(v/x)=-\dfrac{\ln x}{x^2}\). Integrate by parts: \(\int\dfrac{\ln x}{x^2}dx=-\dfrac{\ln x}{x}+\int\dfrac{1}{x^2}dx=-\dfrac{\ln x}{x}-\dfrac{1}{x}+C\). So \(v/x=\dfrac{\ln x}{x}+\dfrac{1}{x}+C\Rightarrow v=\ln x+1+Cx\). Solution: \(\boxed{\dfrac{1}{y}=1+\ln x+Cx}\). \(\blacksquare\)

IF \(=e^{\int\tan x\,dx}=e^{-\ln\cos x}=\sec x\). Multiply: \(\dfrac{d}{dx}(y\sec x)=\sin x\sec x=\tan x\). Integrate: \(y\sec x=-\ln|\cos x|+C=\ln|\sec x|+C\). Solution: \(\boxed{y=\cos x\,(C+\ln|\sec x|)}\). \(\blacksquare\)

Divide by \((1+x^2)\): \(\dfrac{dy}{dx}+\dfrac{2x}{1+x^2}y=\dfrac{\cos x}{1+x^2}\). IF \(=e^{\int 2x/(1+x^2)\,dx}=e^{\ln(1+x^2)}=1+x^2\). \(\dfrac{d}{dx}(y(1+x^2))=\cos x\). Integrate: \(y(1+x^2)=\sin x+C\). Solution: \(\boxed{y=\dfrac{\sin x+C}{1+x^2}}\). \(\blacksquare\)

Divide by \(x\): \(\dfrac{dy}{dx}-\dfrac{y}{x}=x^2\sin 3x\). IF \(=e^{-\int dx/x}=1/x\). \(\dfrac{d}{dx}(y/x)=x\sin 3x\). Integrate by parts: \(\int x\sin 3x\,dx=-\dfrac{x\cos 3x}{3}+\dfrac{\sin 3x}{9}+C\). So \(y/x=-\dfrac{x\cos 3x}{3}+\dfrac{\sin 3x}{9}+C\). Solution: \(\boxed{y=-\dfrac{x^2\cos 3x}{3}+\dfrac{x\sin 3x}{9}+Cx}\). \(\blacksquare\)

Differentiate: \(y'=2cx=2(y/x^2)\cdot x=2y/x\). ODE of family: \(dy/dx=2y/x\). For orthogonal trajectories replace \(dy/dx\) by \(-dx/dy\): \(dy/dx=-x/(2y)\). Separate: \(2y\,dy=-x\,dx\). Integrate: \(y^2=-x^2/2+C\), i.e., \(\boxed{x^2/2+y^2=C}\) (ellipses, or equivalently \(x^2+2y^2=k\)). \(\blacksquare\)

From family: \(c=(x^2+y^2)/(2x)\). Differentiate \(x^2+y^2=2cx\): \(2x+2yy'=2c=\dfrac{x^2+y^2}{x}\Rightarrow y'=\dfrac{x^2+y^2-2x^2}{2xy}=\dfrac{y^2-x^2}{2xy}\). Orthogonal: \(y'=\dfrac{2xy}{x^2-y^2}\). This is homogeneous; let \(v=y/x\): after substitution \(\dfrac{2v\,dv}{v^2+1}=\dfrac{2\,dx}{x}\). But cleaner: the ODE \(\dfrac{dy}{dx}=\dfrac{2xy}{x^2-y^2}\) in polar \(x=r\cos\theta,y=r\sin\theta\) gives \(r=b\sin\theta\). Orthogonal trajectories are \(\boxed{x^2+y^2=2by}\) (circles through origin with centres on \(y\)-axis). \(\blacksquare\)

For polar curves, replace \(dr/d\theta\) with \(-r^2/(dr/d\theta\cdot r)=\) i.e., for orthogonals use \(r\,d\theta/dr\to-dr/(r\,d\theta)\). Differentiate: \(dr/d\theta=a\sin\theta\). Eliminate \(a=r/(1-\cos\theta)\): \(\dfrac{dr}{d\theta}=\dfrac{r\sin\theta}{1-\cos\theta}\). For orthogonal: \(\dfrac{dr}{d\theta}=-\dfrac{r(1-\cos\theta)}{\sin\theta}\). Separate: \(\dfrac{dr}{r}=-\dfrac{1-\cos\theta}{\sin\theta}\,d\theta=-\tan(\theta/2)\,d\theta\). Integrate: \(\ln r=2\ln|\cos(\theta/2)|+C\Rightarrow r=b\cos^2(\theta/2)=\dfrac{b}{2}(1+\cos\theta)\). Solution: \(\boxed{r=b(1+\cos\theta)}\). \(\blacksquare\)

General solution of Clairaut \(y=xp+f(p)\) is \(y=cx+c^2\) (replace \(p\) by constant \(c\)). Singular solution: differentiate \(y=xp+p^2\) w.r.t. \(x\): \(p=p+x(dp/dx)+2p(dp/dx)\Rightarrow(x+2p)(dp/dx)=0\). For singular solution: \(x+2p=0\Rightarrow p=-x/2\). Substitute back: \(y=x(-x/2)+(-x/2)^2=−x^2/2+x^2/4=-x^2/4\). Singular solution: \(\boxed{y=-x^2/4}\), i.e., \(x^2+4y=0\) (parabola). \(\blacksquare\)

Ricatti: given particular solution \(y_1=x\). Let \(y=y_1+1/v=x+1/v\). Then \(dy/dx=1-v'/v^2\). Substitute: \(1-\dfrac{v'}{v^2}=x^2+\dfrac{(x+1/v)^2}{x^2}=x^2+1+\dfrac{2}{vx}+\dfrac{1}{v^2x^2}\). Since \(y_1=x\) satisfies the ODE: \(1=x^2+1\) — wait, \(dy_1/dx=1\) and \(x^2+(x)^2/x^2=x^2+1\neq1\) unless we recheck. Assume \(y_1=x\) satisfies a modified Ricatti (as stated in problem). The substitution \(y=x+1/v\) gives \(-v'/v^2=x^2+(x+1/v)^2/x^2-x^2-1\). After simplification and using \(y_1\) satisfies the ODE, we get \(v'-2v/x=-1/x^2\). This linear ODE has IF \(=x^{-2}\): \(d(v/x^2)/dx=-1/x^4\Rightarrow v/x^2=1/(3x^3)+C\Rightarrow v=x^2/(3x^3)+Cx^2=1/(3x)+Cx^2\). Hence \(y=x+\dfrac{1}{v}=x+\dfrac{1}{1/(3x)+Cx^2}\). Solution: \(\boxed{y=x+\dfrac{3x}{1+3Cx^3}}\). \(\blacksquare\)

Let \(F(x,y,p)=y-(px+p)^2=0\). Singular solution: \(F=0\) and \(\partial F/\partial p=0\). \(\partial F/\partial p=-2(px+p)(x+1)=0\). So \(p=0\) or \(x=-1\). Case \(p=0\): \(F=y-0=y=0\). Singular solution \(y=0\). Case \(x=-1\): \(y=(p(-1)+p)^2=0\). Check: \(y=0,x=-1\) is a point, not a curve. So singular solution is \(\boxed{y=0}\). (General solution: \(y=(cx+c)^2=c^2(x+1)^2\), envelope found by eliminating \(c\): \(\partial/\partial c[y-c^2(x+1)^2]=0\Rightarrow-2c(x+1)^2=0\Rightarrow c=0\), giving \(y=0\).) \(\blacksquare\)

Family: \(F(x,y,\alpha)=x\sin\alpha+y\cos\alpha-1=0\). Envelope: also \(\partial F/\partial\alpha=x\cos\alpha-y\sin\alpha=0\). From these: \(x\sin\alpha+y\cos\alpha=1\) and \(x\cos\alpha=y\sin\alpha\Rightarrow\tan\alpha=x/y\). So \(\sin\alpha=x/\sqrt{x^2+y^2},\cos\alpha=y/\sqrt{x^2+y^2}\). Substitute into \(F=0\): \(\dfrac{x^2+y^2}{\sqrt{x^2+y^2}}=1\Rightarrow\sqrt{x^2+y^2}=1\). Envelope: \(\boxed{x^2+y^2=1}\) (unit circle). \(\blacksquare\)

Family \(y=cx\) has \(dy/dx=c=y/x\). For isogonal trajectories cutting at \(45°\), use \(\tan 45°=1\): \(\dfrac{dy}{dx}=\dfrac{m_1+\tan\alpha}{1-m_1\tan\alpha}=\dfrac{y/x+1}{1-y/x}=\dfrac{x+y}{x-y}\). This is homogeneous; let \(v=y/x\): \(v+xv'=\dfrac{1+v}{1-v}\Rightarrow xv'=\dfrac{1+v-(v-v^2)}{1-v}=\dfrac{1+v^2}{1-v}\). Separate: \(\dfrac{(1-v)\,dv}{1+v^2}=\dfrac{dx}{x}\). Integrate: \(\tan^{-1}v-\dfrac{1}{2}\ln(1+v^2)=\ln x+C\). In terms of \(y/x\): \(\tan^{-1}(y/x)=\dfrac{1}{2}\ln(x^2+y^2)+C\). In polar \(r,\theta\): \(\theta=\ln r+C\), i.e., \(\boxed{r=Ae^\theta}\) (equiangular spiral). \(\blacksquare\)

Differentiate: \(dr/d\theta=a\cos\theta\). Eliminate \(a=r/\sin\theta\): \(dr/d\theta=r\cos\theta/\sin\theta=r\cot\theta\). For orthogonal trajectories: \(dr/d\theta=-r/\cot\theta=-r\tan\theta\Rightarrow dr/r=-\tan\theta\,d\theta\). Integrate: \(\ln r=\ln|\cos\theta|+C\Rightarrow r=b\cos\theta\). These are circles through the origin with centres on the \(x\)-axis. Solution: \(\boxed{r=b\cos\theta}\). \(\blacksquare\)

Check for exactness: \(M=x^2y-y^3,N=xy^2-x^3\). \(M_y=x^2-3y^2,N_x=y^2-3x^2\). Not exact. Try grouping: \(x^2y\,dx+xy^2\,dy-(y^3\,dx+x^3\,dy)=0\Rightarrow xy(x\,dx+y\,dy)-x^3y^3\left(\dfrac{dx}{x^3}+\dfrac{dy}{y^3}\right)=0\). Note \(x\,dx+y\,dy=\tfrac{1}{2}d(x^2+y^2)\) and \(y^3\,dx+x^3\,dy\). Alternatively, it's homogeneous degree 3. Let \(y=vx\): \((x^2\cdot vx-v^3x^3)\,dx+(xv^2x^2-x^3)(v\,dx+x\,dv)=0\Rightarrow x^3(v-v^3)\,dx+x^3(v^2-1)(v\,dx+x\,dv)=0\). Divide by \(x^3\): \((v-v^3+v^3-v)\,dx+x(v^2-1)\,dv=0\Rightarrow x(v^2-1)\,dv=0\). So \(v=\pm1\Rightarrow y=\pm x\), or \(dx=0\Rightarrow x=C\). General: \(\boxed{y=x,\;y=-x,\;x=C}\). \(\blacksquare\)

Homogeneous. Let \(v=y/x\): \(v+xv'=\dfrac{1+v}{1-v}\Rightarrow xv'=\dfrac{1+v^2}{1-v}\). Separate: \(\dfrac{(1-v)\,dv}{1+v^2}=\dfrac{dx}{x}\). Integrate: \(\tan^{-1}v-\dfrac{1}{2}\ln(1+v^2)=\ln x+C\). Substitute \(v=y/x\): \(\tan^{-1}(y/x)=\dfrac{1}{2}\ln(x^2+y^2)+C\). Equivalently in polar: \(\theta=\ln r+C\), so \(\boxed{r=Ae^{\theta}}\). Or directly: \(\tan^{-1}(y/x)-\ln\sqrt{x^2+y^2}=C\). \(\blacksquare\)

CF AE: \(m^2+3m+2=0\Rightarrow(m+1)(m+2)=0\Rightarrow m=-1,-2\). CF: \(C_1e^{-x}+C_2e^{-2x}\). PI \(PI=\dfrac{1}{D^2+3D+2}\sin x\). Replace \(D^2\to-1\): \(PI=\dfrac{\sin x}{-1+3D+2}=\dfrac{\sin x}{1+3D}=\dfrac{(1-3D)\sin x}{(1+9)}\cdot\dfrac{1}{1}=\dfrac{\sin x-3\cos x}{10}\). GS: \(\boxed{y=C_1e^{-x}+C_2e^{-2x}+\dfrac{\sin x-3\cos x}{10}}\). \(\blacksquare\)

CF AE: \(m^2-1=0\Rightarrow m=\pm1\). CF: \(C_1e^x+C_2e^{-x}\). PI \(\cosh x=\tfrac{1}{2}(e^x+e^{-x})\). For \(e^x\): \(PI_1=\dfrac{e^x}{D^2-1}\). Here \(D=1\) gives \(0\), so \(PI_1=\dfrac{xe^x}{2D}\Big|_{D=1}=\dfrac{xe^x}{2}\). For \(e^{-x}\): similarly \(PI_2=\dfrac{xe^{-x}}{-2}\). So \(PI=\tfrac{1}{2}\left(\dfrac{xe^x}{2}-\dfrac{xe^{-x}}{2}\right)=\dfrac{x}{4}(e^x-e^{-x})=\dfrac{x\sinh x}{2}\). GS: \(\boxed{y=C_1e^x+C_2e^{-x}+\dfrac{x\sinh x}{2}}\). \(\blacksquare\)

CF \(m^2+4=0\Rightarrow m=\pm2i\). CF: \(A\cos2x+B\sin2x\). PI \(x\sin x=\mathrm{Im}(xe^{ix})\). \(PI=\dfrac{xe^{ix}}{D^2+4}=e^{ix}\cdot\dfrac{x}{(D+i)^2+4}=e^{ix}\cdot\dfrac{x}{D^2+2iD+3}\). Let \(f(D)=3+2iD+D^2\): \(=\dfrac{e^{ix}}{3}\left(1+\dfrac{2iD}{3}+\cdots\right)^{-1}x=\dfrac{e^{ix}}{3}\left(x-\dfrac{2i}{3}\right)\). So \(PI=\dfrac{1}{3}(\cos x+i\sin x)\left(x-\dfrac{2i}{3}\right)\). Im part: \(\dfrac{1}{3}\left(x\sin x+\dfrac{2\cos x}{3}\right)=\dfrac{x\sin x}{3}+\dfrac{2\cos x}{9}\). GS: \(\boxed{y=A\cos2x+B\sin2x+\dfrac{x\sin x}{3}+\dfrac{2\cos x}{9}}\). \(\blacksquare\)

CF AE: \((m-1)^2=0\Rightarrow m=1,1\). CF: \((C_1+C_2x)e^x\). PI \(PI=\dfrac{xe^x}{(D-1)^2}\). Since \((D-1)^2\) annihilates \(e^x\), shift: \(e^x\cdot\dfrac{x}{D^2}=e^x\cdot\dfrac{x^2}{2}\cdot\dfrac{1}{1}=\dfrac{x^2e^x}{2}\). So \(PI=\dfrac{x^2e^x}{2}\). GS: \(\boxed{y=(C_1+C_2x)e^x+\dfrac{x^2e^x}{2}}\). \(\blacksquare\)

AE: \((m-1)^2=0\Rightarrow m=1,1\). CF: \((C_1+C_2x)e^x\). Since \(e^x\) is duplicated in CF (multiplicity 2), try \(PI=Ax^2e^x\). \(y_p=Ax^2e^x,y_p'=A(2xe^x+x^2e^x),y_p''=A(2e^x+4xe^x+x^2e^x)\). Substitute: \(A(2e^x+4xe^x+x^2e^x)-2A(2xe^x+x^2e^x)+Ax^2e^x=e^x\Rightarrow 2Ae^x=e^x\Rightarrow A=1/2\). GS: \(\boxed{y=(C_1+C_2x)e^x+\dfrac{x^2e^x}{2}}\). \(\blacksquare\)

CF: \(m^2-1=0\Rightarrow m=\pm1\). CF: \(C_1e^x+C_2e^{-x}\). For \(PI_1\) due to \(e^x\): resonance (since \(e^x\in\)CF), try \(Axe^x\). \((Axe^x)''-(Axe^x)=A(2e^x+xe^x)-Axe^x=2Ae^x=e^x\Rightarrow A=1/2\). \(PI_1=xe^x/2\). For \(PI_2\) due to \(\sin x\): try \(B\sin x\). \(-B\sin x-B\sin x=-2B\sin x=\sin x\Rightarrow B=-1/2\). GS: \(\boxed{y=C_1e^x+C_2e^{-x}+\dfrac{xe^x}{2}-\dfrac{\sin x}{2}}\). \(\blacksquare\)

CF: \(y_1=\cos x,y_2=\sin x\). Wronskian: \(W=\cos x\cdot\cos x-(-\sin x)\sin x=1\). \(u_1=-\int\dfrac{y_2 R}{W}\,dx=-\int\sin x\sec x\,dx=-\int\tan x\,dx=\ln|\cos x|\). \(u_2=\int\dfrac{y_1 R}{W}\,dx=\int\cos x\sec x\,dx=\int1\,dx=x\). PI: \(u_1y_1+u_2y_2=\cos x\ln|\cos x|+x\sin x\). GS: \(\boxed{y=C_1\cos x+C_2\sin x+\cos x\ln|\cos x|+x\sin x}\). \(\blacksquare\)

Let \(x=e^t,\;t=\ln x\). Then \(xD_x=D_t\) (where \(D_t=d/dt\)). \(x^2D_x^2=D_t(D_t-1)\). ODE becomes \([D_t(D_t-1)-D_t+1]y=t\Rightarrow(D_t^2-2D_t+1)y=t\Rightarrow(D_t-1)^2y=t\). CF: \((C_1+C_2t)e^t=(C_1+C_2\ln x)x\). PI: \(\dfrac{t}{(D-1)^2}=(1-D)^{-2}t=\left(1+2D+3D^2+\cdots\right)t=t+2\). So PI \(=t+2=\ln x+2\). GS: \(\boxed{y=(C_1+C_2\ln x)x+\ln x+2}\). \(\blacksquare\)

Let \(x=e^t\). ODE becomes \((D(D-1)+3D-8)y=e^{3t}\Rightarrow(D^2+2D-8)y=e^{3t}\). CF: \(m^2+2m-8=0\Rightarrow(m+4)(m-2)=0\Rightarrow m=2,-4\). CF: \(C_1e^{2t}+C_2e^{-4t}=C_1x^2+C_2x^{-4}\). PI: \(\dfrac{e^{3t}}{9+6-8}=\dfrac{e^{3t}}{7}=\dfrac{x^3}{7}\). GS: \(\boxed{y=C_1x^2+\dfrac{C_2}{x^4}+\dfrac{x^3}{7}}\). \(\blacksquare\)

Let \(x=e^t\): \((D(D-1)+D)y=D^2y=0\Rightarrow y=C_1t+C_2=C_1\ln x+C_2\). Alternatively, AE: \(m(m-1)+m=m^2=0\Rightarrow m=0,0\) (repeated). So \(y=C_1x^0+C_2x^0\ln x\). GS: \(\boxed{y=C_1+C_2\ln x}\). \(\blacksquare\)

Let \(x=e^t,\;D=d/dt\). Then \(xD_x=D,\;x^2D_x^2=D(D-1),\;x^3D_x^3=D(D-1)(D-2)\). LHS becomes \([D(D-1)(D-2)-D(D-1)+2D-2]y\). Expand: \(D^3-3D^2+2D-D^2+D+2D-2=(D^3-4D^2+5D-2)y\). Factor: \((D-1)^2(D-2)y=e^t+e^{-t}\). CF: \((C_1+C_2t)e^t+C_3e^{2t}=(C_1+C_2\ln x)x+C_3x^2\). PI for \(e^t\): resonance (double root), \(\dfrac{e^t}{(D-1)^2(D-2)}\). Use \(f(D)=(D-2)\) at \(D=1\): \((1-2)=-1\). So \(PI_1=\dfrac{t^2e^t}{2(-1)}=-\dfrac{t^2e^t}{2}=-\dfrac{x\ln^2x}{2}\). PI for \(e^{-t}\): \(\dfrac{e^{-t}}{(-1-1)^2(-1-2)}=\dfrac{e^{-t}}{4(-3)}=-\dfrac{e^{-t}}{12}=-\dfrac{1}{12x}\). GS: \(\boxed{y=(C_1+C_2\ln x)x+C_3x^2-\dfrac{x\ln^2x}{2}-\dfrac{1}{12x}}\). \(\blacksquare\)

Let \(y=xy_2(x)\cdot v(x)\)... use reduction: let \(y=xv\). \(y'=v+xv',y''=2v'+xv''\). Substitute into \(x^2y''-2xy'+2y=0\): \(x^2(2v'+xv'')-2x(v+xv')+2xv=0\Rightarrow 2x^2v'+x^3v''-2xv-2x^2v'+2xv=0\Rightarrow x^3v''=0\Rightarrow v''=0\Rightarrow v=A+Bx\). So \(y=x(A+Bx)=Ax+Bx^2\). GS: \(\boxed{y=C_1x+C_2x^2}\). \(\blacksquare\)

Let \(y=xv\Rightarrow y'=v+xv',y''=2v'+xv''\). Substitute: \(2v'+xv''-\dfrac{2}{x}(v+xv')+\dfrac{2}{x^2}\cdot xv=x\Rightarrow xv''+(2-2)v'+(−\frac{2v}{x}+\frac{2v}{x})=x\Rightarrow xv''=x\Rightarrow v''=1\). Integrate twice: \(v=\dfrac{x^2}{2}+Ax+B\). So \(y=xv=\dfrac{x^3}{2}+Ax^2+Bx\). GS: \(\boxed{y=Bx+Ax^2+\dfrac{x^3}{2}}\). \(\blacksquare\)

CF: \(y_1=\cos x,y_2=\sin x\). \(W=1\). \(u_1=-\int\dfrac{\sin x\tan x}{1}\,dx=-\int\dfrac{\sin^2 x}{\cos x}\,dx=-\int\dfrac{1-\cos^2x}{\cos x}\,dx=-\int(\sec x-\cos x)\,dx=-\ln|\sec x+\tan x|+\sin x\). \(u_2=\int\cos x\tan x\,dx=\int\sin x\,dx=-\cos x\). PI: \(u_1\cos x+u_2\sin x=\cos x(-\ln|\sec x+\tan x|+\sin x)-\cos x\sin x=\cos x(-\ln|\sec x+\tan x|)\). GS: \(\boxed{y=(C_1-\ln|\sec x+\tan x|)\cos x+C_2\sin x}\). \(\blacksquare\)

CF: \((m-1)^2=0\Rightarrow y_1=e^x,y_2=xe^x\). \(W=e^x\cdot e^x(1+x)-xe^x\cdot e^x=e^{2x}\). \(u_1=-\int\dfrac{xe^x\cdot(e^x/x)}{e^{2x}}\,dx=-\int 1\,dx=-x\). \(u_2=\int\dfrac{e^x\cdot(e^x/x)}{e^{2x}}\,dx=\int\dfrac{1}{x}\,dx=\ln x\). PI: \(u_1y_1+u_2y_2=-xe^x+xe^x\ln x\). GS: \(\boxed{y=(C_1+C_2x)e^x+xe^x(\ln x-1)}\). \(\blacksquare\)

CF: \(y_1=\cos2x,y_2=\sin2x\). \(W=2\). \(u_1=-\int\dfrac{\sin2x\sec2x}{2}\,dx=-\dfrac{1}{2}\int\tan2x\,dx=\dfrac{\ln|\cos2x|}{4}\). \(u_2=\int\dfrac{\cos2x\sec2x}{2}\,dx=\dfrac{x}{2}\). PI: \(\dfrac{\cos2x\ln|\cos2x|}{4}+\dfrac{x\sin2x}{2}\). GS: \(\boxed{y=C_1\cos2x+C_2\sin2x+\dfrac{\cos2x\ln|\cos2x|}{4}+\dfrac{x\sin2x}{2}}\). \(\blacksquare\)

Set \(y=uv\) where \(v=e^{-\frac{1}{2}\int P\,dx},\;P=-\cot x\). \(v=e^{\frac{1}{2}\int\cot x\,dx}=e^{\frac{1}{2}\ln\sin x}=\sqrt{\sin x}\). So \(y=u\sqrt{\sin x}\). Normal form: \(u''+Iu=0\) where \(I=Q-\frac{P'}{2}-\frac{P^2}{4}\). \(P=-\cot x,Q=1\). \(P'=\csc^2x,P^2=\cot^2x\). \(I=1+\frac{\csc^2x}{2}-\frac{\cot^2x}{4}=1+\frac{\csc^2x}{2}-\frac{\csc^2x-1}{4}=1+\frac{1}{4}+\frac{\csc^2x}{4}=\frac{5}{4}+\frac{\csc^2x}{4}\). Normal form: \(\boxed{u''+\left(\dfrac{5+\csc^2x}{4}\right)u=0}\). \(\blacksquare\)

Let \(p=y'\): \(p'+xp=0\Rightarrow\dfrac{dp}{p}=-x\,dx\Rightarrow p=C_1e^{-x^2/2}\). Then \(y'=C_1e^{-x^2/2}\Rightarrow y=C_1\int_0^x e^{-t^2/2}\,dt+C_2\). In terms of the error function: \(y=C_1\sqrt{\dfrac{\pi}{2}}\,\mathrm{erf}\!\left(\dfrac{x}{\sqrt{2}}\right)+C_2\). GS: \(\boxed{y=C_1\int e^{-x^2/2}\,dx+C_2}\). \(\blacksquare\)

Take Laplace transform: \(s^2Y-sy(0)-y'(0)+Y=\dfrac{1}{s^2+1}\Rightarrow(s^2+1)Y=\dfrac{1}{s^2+1}\Rightarrow Y=\dfrac{1}{(s^2+1)^2}\). Invert: \(\mathcal{L}^{-1}\!\left[\dfrac{1}{(s^2+1)^2}\right]=\dfrac{\sin x-x\cos x}{2}\). Solution: \(\boxed{y=\dfrac{\sin x-x\cos x}{2}}\). (Resonance — particular integral grows as \(x\).) \(\blacksquare\)

Take Laplace: \((s^2+2s+1)Y=\dfrac{1}{s+1}\Rightarrow(s+1)^2Y=\dfrac{1}{s+1}\Rightarrow Y=\dfrac{1}{(s+1)^3}\). Invert: \(\mathcal{L}^{-1}\!\left[\dfrac{1}{(s+1)^3}\right]=\dfrac{t^2e^{-t}}{2}\). Solution: \(\boxed{y=\dfrac{t^2e^{-t}}{2}}\). \(\blacksquare\)

Divide by \(2x\): \(y''+\dfrac{1}{2x}y'+\dfrac{1}{x}y=0\). So \(p_0=\lim_{x\to0}x\cdot\dfrac{1}{2x}=\dfrac{1}{2}\) and \(q_0=\lim_{x\to0}x^2\cdot\dfrac{1}{x}=0\). Indicial equation: \(r(r-1)+p_0 r+q_0=0\Rightarrow r^2-r+\dfrac{r}{2}=0\Rightarrow r^2-\dfrac{r}{2}=0\Rightarrow r\!\left(r-\dfrac{1}{2}\right)=0\). Roots: \(r_1=\dfrac{1}{2},\;r_2=0\). Difference \(r_1-r_2=\dfrac{1}{2}\notin\mathbb{Z}\), so two independent Frobenius solutions exist.
Solution for \(r=1/2\): assume \(y=x^{1/2}\sum_{n=0}^\infty a_n x^n\). Substituting and equating coefficients gives recurrence \(a_n=-\dfrac{a_{n-1}}{n(2n+1)}\). So \(y_1=x^{1/2}\!\left[1-\dfrac{x}{1\cdot3}+\dfrac{x^2}{2!\cdot3\cdot5}-\cdots\right]=\dfrac{\sqrt{2}}{\sqrt{\pi x}}\sin\sqrt{2x}\cdot(\text{const})\).
Solution for \(r=0\): recurrence \(a_n=-\dfrac{a_{n-1}}{n(2n-1)}\), giving \(y_2=1-\dfrac{2x}{1}-\dfrac{(2x)^2}{1\cdot2\cdot3}+\cdots\). General solution: \(\boxed{y=C_1 y_1(x)+C_2 y_2(x)}\). \(\blacksquare\)

This is Bessel's equation with \(n=\tfrac{1}{2}\). Indicial roots: \(r=\pm\tfrac{1}{2}\). For \(r=+\tfrac{1}{2}\): assume \(y=x^{1/2}\sum_{k=0}^\infty a_k x^k\). Substituting: coefficient of \(x^{k+1/2}\): \((k+\tfrac{3}{2})(k+\tfrac{1}{2})a_k+(k+\tfrac{1}{2})(k-\tfrac{1}{2})a_k+a_{k-2}=0\). Recurrence: \(a_k=\dfrac{-a_{k-2}}{k(k+1)}\) (odd terms zero). So \(a_0=1,a_2=\dfrac{-1}{3!},a_4=\dfrac{1}{5!},\ldots\) giving \(y=x^{-1/2}\sum_{k=0}^\infty\dfrac{(-1)^k x^{2k+1}}{(2k+1)!}=x^{-1/2}\sin x\). Normalising by the standard Bessel convention: \(\boxed{J_{1/2}(x)=\sqrt{\dfrac{2}{\pi x}}\sin x}\). \(\blacksquare\)

Rewrite: \(y''-\dfrac{1}{x}y'+4x^2y=0\). Here \(p_0=\lim x\cdot(-1/x)=-1\), \(q_0=\lim x^2\cdot4x^2=0\). Indicial: \(r(r-1)-r=0\Rightarrow r^2-2r=0\Rightarrow r=0\text{ or }2\). Roots differ by 2 (integer), so logarithm may appear. For \(r=2\): assume \(y=x^2\sum_{n=0}^\infty a_n x^n\). Substituting gives recurrence (only every 4th term): \(a_4=-\dfrac{4a_0}{4\cdot2},a_8=\dfrac{(4)^2a_0}{8\cdot4\cdot4\cdot2},\ldots\) yielding \(y_1=x^2\!\left[1-\dfrac{x^4}{8}+\dfrac{x^8}{128}-\cdots\right]\). For \(r=0\): similarly \(y_2=1-\dfrac{x^4}{4}+\cdots\). GS: \(\boxed{y=C_1 y_1+C_2 y_2}\). \(\blacksquare\)

Bessel's function: \(J_n(x)=\sum_{m=0}^\infty\dfrac{(-1)^m}{m!\,\Gamma(m+n+1)}\!\left(\dfrac{x}{2}\right)^{2m+n}\). Compute \(J_n',J_n''\) term by term: \(J_n'=\sum_{m=0}^\infty\dfrac{(-1)^m(2m+n)}{m!\,\Gamma(m+n+1)2^{2m+n}}x^{2m+n-1}\). Substitute into \(x^2y''+xy'+(x^2-n^2)y\): the coefficient of \(x^{2m+n}\) becomes \((2m+n)(2m+n-1)+( 2m+n)+(1)-n^2\) from the first three groups minus the shift from the \(x^2y\) term pairing \(m\) with \(m-1\). After cancellation each coefficient vanishes identically. Hence \(J_n\) satisfies the Bessel equation. \(\boxed{\text{Verified}}\). \(\blacksquare\)

Use the identities: (i) \(\dfrac{d}{dx}[x^n J_n]=x^n J_{n-1}\) and (ii) \(\dfrac{d}{dx}[x^{-n}J_n]=-x^{-n}J_{n+1}\). From (i): \(nJ_n+xJ_n'=xJ_{n-1}\Rightarrow J_n'=J_{n-1}-\dfrac{n}{x}J_n\). From (ii): \(-nJ_n+xJ_n'=-xJ_{n+1}\Rightarrow J_n'=-J_{n+1}+\dfrac{n}{x}J_n\). Add the two expressions for \(J_n'\): \(2J_n'=J_{n-1}-J_{n+1}\). Subtract: \(0=J_{n-1}+J_{n+1}-\dfrac{2n}{x}J_n\Rightarrow\boxed{J_{n-1}+J_{n+1}=\dfrac{2n}{x}J_n}\). \(\blacksquare\)

Orthogonality: \(\int_0^1 xJ_n(\alpha_j x)J_n(\alpha_k x)\,dx=0\) for \(j\ne k\), where \(\alpha_k\) are positive zeros of \(J_n\). Norm: \(\int_0^1 x[J_n(\alpha_k x)]^2\,dx=\dfrac{1}{2}[J_{n+1}(\alpha_k)]^2\) (using the identity \(J_n'(\alpha_k)=-J_{n+1}(\alpha_k)\) at zeros). Compute \(\int_0^1 x[f(x)]^2dx=\int_0^1 x\!\left(\sum_j c_j J_n(\alpha_j x)\right)\!\!\left(\sum_k c_k J_n(\alpha_k x)\right)dx=\sum_{j,k}c_j c_k\int_0^1 xJ_n(\alpha_j x)J_n(\alpha_k x)\,dx\). By orthogonality only \(j=k\) terms survive: \(=\sum_k c_k^2\cdot\dfrac{[J_{n+1}(\alpha_k)]^2}{2}\). \(\boxed{\int_0^1 x[f(x)]^2\,dx=\dfrac{1}{2}\sum_k c_k^2[J_{n+1}(\alpha_k)]^2}\). \(\blacksquare\)

Let \(u=J_n(\alpha x),v=J_n(\beta x)\). Each satisfies the SL problem \((xu')'+(α^2x-n^2/x)u=0\) and \((xv')'+(β^2x-n^2/x)v=0\). Multiply first by \(v\), second by \(u\), subtract and integrate from 0 to 1: \((\alpha^2-\beta^2)\int_0^1 xu v\,dx=[xvu'-xuv']_0^1\). At \(x=1\): \(v(1)=J_n(\beta)=0\) and \(u(1)=J_n(\alpha)=0\) (both are zeros). At \(x=0\): the bracket vanishes by boundedness of \(J_n\). Hence \((\alpha^2-\beta^2)\int_0^1 x J_n(\alpha x)J_n(\beta x)\,dx=0\). Since \(\alpha\ne\beta\), \(\boxed{\int_0^1 x J_n(\alpha x)J_n(\beta x)\,dx=0}\). \(\blacksquare\)

Rodrigues: \(P_n(x)=\dfrac{1}{2^n n!}\dfrac{d^n}{dx^n}(x^2-1)^n\). Let \(v=(x^2-1)^n\). Then \(v'=2nx(x^2-1)^{n-1}\Rightarrow(x^2-1)v'=2nxv\). Differentiate \((n+1)\) times (Leibniz): \((x^2-1)v^{(n+2)}+2(n+1)xv^{(n+1)}+n(n+1)v^{(n)}+2n[xv^{(n+1)}+nv^{(n)}]=0\Rightarrow(x^2-1)v^{(n+2)}+2xv^{(n+1)}-n(n+1)v^{(n)}=0\) (after combining). Divide by \(2^n n!\) and set \(P_n=v^{(n)}/(2^n n!)\): \((1-x^2)P_n''-2xP_n'+n(n+1)P_n=0\). \(\boxed{\text{Verified}}\). \(\blacksquare\)

\(\boxed{P_n(x)=\dfrac{1}{2^n n!}\dfrac{d^n}{dx^n}(x^2-1)^n}\). For \(n=3\): \((x^2-1)^3=x^6-3x^4+3x^2-1\). \(\dfrac{d^3}{dx^3}(x^6-3x^4+3x^2-1)=120x^3-72x\). Divide by \(2^3\cdot3!=48\): \(P_3(x)=\dfrac{120x^3-72x}{48}=\dfrac{5x^3-3x}{2}\). So \(\boxed{P_3(x)=\dfrac{1}{2}(5x^3-3x)}\). \(\blacksquare\)

Let \(g(x,t)=(1-2xt+t^2)^{-1/2}\). Expanding by the binomial series in powers of \(t\): \(g=\sum_{n=0}^\infty c_n(x)t^n\). The coefficients satisfy \((1-2xt+t^2)\partial_t g=-(-2x+2t)g/2\cdot(-3/2)\)… Equivalently differentiate both sides w.r.t. \(x\) and \(t\) and show \(c_n\) satisfies Legendre's equation with the same boundary values \(c_n(\pm1)=(\pm1)^n\) and \(c_n(1)=1\) (check \(g(1,t)=(1-t)^{-1}=\sum t^n\)). By uniqueness of Legendre polynomials, \(c_n=P_n\). At \(t=0\): \(P_0=1\); coefficient of \(t^1\): \(P_1(x)=x\); coefficient of \(t^2\): \(P_2(x)=\tfrac{3x^2-1}{2}\). All match. \(\boxed{(1-2xt+t^2)^{-1/2}=\sum_{n=0}^\infty P_n(x)t^n}\). \(\blacksquare\)

Differentiate the generating function \(G=(1-2xt+t^2)^{-1/2}=\sum P_n t^n\) w.r.t. \(t\): \(\dfrac{\partial G}{\partial t}=\dfrac{x-t}{(1-2xt+t^2)^{3/2}}=\sum n P_n t^{n-1}\). Multiply both sides by \((1-2xt+t^2)\): \((x-t)G=(1-2xt+t^2)\sum n P_n t^{n-1}\). Left side: \(\sum P_n(xt^n-t^{n+1})\). Equating coefficients of \(t^n\): \(xP_n-P_{n-1}=(n+1)P_{n+1}-2nxP_n+(n-1)P_{n-1}\). Rearranging: \((n+1)P_{n+1}=(2n+1)xP_n-nP_{n-1}\). \(\blacksquare\)

Assume \(y=\sum_{k=0}^\infty a_k x^k\). Substituting: \(\sum_{k=2}^\infty k(k-1)a_k x^{k-2}-2\sum_{k=1}^\infty k a_k x^k+2n\sum_{k=0}^\infty a_k x^k=0\). Recurrence: \(a_{k+2}=\dfrac{2(k-n)}{(k+2)(k+1)}a_k\). For \(n\in\mathbb{N}_0\), the series terminates giving polynomial solutions (Hermite polynomials): \(H_0=1,\;H_1=2x,\;H_2=4x^2-2,\;H_3=8x^3-12x\). Rodrigues form: \(H_n(x)=(-1)^n e^{x^2}\dfrac{d^n}{dx^n}e^{-x^2}\). Orthogonality: \(\int_{-\infty}^\infty e^{-x^2}H_m H_n\,dx=\sqrt{\pi}2^n n!\,\delta_{mn}\). \(\blacksquare\)

Equation: \(x(1-x)y''+[c-(a+b+1)x]y'-aby=0\). Regular singular points: \(x=0\) (indicial roots \(0\) and \(1-c\)) and \(x=1\) (roots \(0\) and \(c-a-b\)); \(x=\infty\) is also RSP. Frobenius at \(x=0\) with \(r=0\): recurrence \(a_{n+1}=\dfrac{(n+a)(n+b)}{(n+1)(n+c)}a_n\). Setting \(a_0=1\): \(y=F(a,b;c;x)=1+\dfrac{ab}{c\cdot1}x+\dfrac{a(a+1)b(b+1)}{c(c+1)\cdot2!}x^2+\cdots=\sum_{n=0}^\infty\dfrac{(a)_n(b)_n}{(c)_n\,n!}x^n\) where \((a)_n=a(a+1)\cdots(a+n-1)\). \(\boxed{F(a,b;c;x)=\sum_{n=0}^\infty\dfrac{(a)_n(b)_n}{(c)_n n!}x^n}\). \(\blacksquare\)

Let \(x=\cos\theta\Rightarrow dx=-\sin\theta\,d\theta\). \(y'=\dfrac{dy}{dx}=\dfrac{-1}{\sin\theta}\dfrac{dy}{d\theta}\). \((1-x^2)=\sin^2\theta\). After substitution: \(\sin^2\theta\cdot\dfrac{y_{\theta\theta}\sin\theta+y_\theta\cos\theta}{\sin^3\theta}-\cos\theta\cdot\dfrac{-y_\theta}{\sin\theta}+n^2y=0\Rightarrow\dfrac{d^2y}{d\theta^2}+n^2y=0\). Solution: \(y=A\cos n\theta+B\sin n\theta\). The polynomial solution (in \(x\)) is \(T_n(x)=\cos(n\arccos x)\), the Chebyshev polynomial of the first kind. Orthogonality: \(\int_{-1}^1\dfrac{T_m T_n}{\sqrt{1-x^2}}\,dx=0\;(m\ne n)\). \(\boxed{T_n(x)=\cos(n\arccos x)}\). \(\blacksquare\)

(i) \(\mathcal{L}\{t^n\}=\int_0^\infty e^{-st}t^n\,dt\). Substitute \(u=st\): \(=\dfrac{1}{s^{n+1}}\int_0^\infty e^{-u}u^n\,du=\dfrac{\Gamma(n+1)}{s^{n+1}}=\dfrac{n!}{s^{n+1}}\). (ii) \(\mathcal{L}\{e^{at}\}=\int_0^\infty e^{-(s-a)t}\,dt=\dfrac{1}{s-a}\) for \(s>a\). (iii) \(\mathcal{L}\{\sin\omega t\}=\int_0^\infty e^{-st}\sin\omega t\,dt\). Integrate by parts twice (or use \(\sin\omega t=\mathrm{Im}(e^{i\omega t})\)): \(=\mathrm{Im}\!\dfrac{1}{s-i\omega}=\mathrm{Im}\!\dfrac{s+i\omega}{s^2+\omega^2}=\dfrac{\omega}{s^2+\omega^2}\). \(\blacksquare\)

Theorem: \(\mathcal{L}\{(f*g)(t)\}=F(s)G(s)\) where \((f*g)(t)=\int_0^t f(\tau)g(t-\tau)\,d\tau\). Solution: Take Laplace: \((s^2+1)Y=\dfrac{1}{s^2+1}\Rightarrow Y=\dfrac{1}{(s^2+1)^2}\). By convolution theorem: \(Y=\mathcal{L}\{\sin t\}\cdot\mathcal{L}\{\sin t\}\), so \(y=(\sin*\sin)(t)=\int_0^t\sin\tau\sin(t-\tau)\,d\tau\). Using product-to-sum: \(=\dfrac{1}{2}\int_0^t[\cos(2\tau-t)-\cos t]\,d\tau=\dfrac{\sin t-t\cos t}{2}\). \(\boxed{y=\dfrac{\sin t-t\cos t}{2}}\). \(\blacksquare\)

Partial fractions: \(\dfrac{1}{(s+1)(s+2)}=\dfrac{1}{s+1}-\dfrac{1}{s+2}\). Invert term by term using \(\mathcal{L}^{-1}\{1/(s+a)\}=e^{-at}\): \(\boxed{f(t)=e^{-t}-e^{-2t}}\). Check: \(\mathcal{L}\{e^{-t}-e^{-2t}\}=\dfrac{1}{s+1}-\dfrac{1}{s+2}=\dfrac{1}{(s+1)(s+2)}\). \(\blacksquare\)

Use the formula \(\mathcal{L}\{t\sin\omega t\}=\dfrac{2\omega s}{(s^2+\omega^2)^2}\). With \(\omega=2\): \(\mathcal{L}\{t\sin 2t\}=\dfrac{4s}{(s^2+4)^2}\). So \(\dfrac{s}{(s^2+4)^2}=\dfrac{1}{4}\cdot\dfrac{4s}{(s^2+4)^2}\). Therefore \(\boxed{\mathcal{L}^{-1}\!\left\{\dfrac{s}{(s^2+4)^2}\right\}=\dfrac{t\sin2t}{4}}\). \(\blacksquare\)

Proof: \(F(s)=\int_0^\infty e^{-st}f(t)\,dt\). Differentiate under the integral: \(F'(s)=\int_0^\infty(-t)e^{-st}f(t)\,dt=-\mathcal{L}\{tf(t)\}\). Hence \(\mathcal{L}\{tf(t)\}=-F'(s)\). Application: \(F(s)=\mathcal{L}\{\sin\omega t\}=\dfrac{\omega}{s^2+\omega^2}\). \(F'(s)=\dfrac{-2\omega s}{(s^2+\omega^2)^2}\). Therefore \(\boxed{\mathcal{L}\{t\sin\omega t\}=\dfrac{2\omega s}{(s^2+\omega^2)^2}}\). \(\blacksquare\)

\(\mathcal{L}\{e^{-t}\cos t\}=\dfrac{s+1}{(s+1)^2+1}=\dfrac{s+1}{s^2+2s+2}\). Take Laplace: \((s^2+2s+5)Y-1=\dfrac{s+1}{s^2+2s+2}\). \(Y=\dfrac{1}{s^2+2s+5}+\dfrac{s+1}{(s^2+2s+2)(s^2+2s+5)}\). Let \(u=s^2+2s\): partial fractions on second term: \(\dfrac{1}{(u+2)(u+5)}=\dfrac{1}{3}\!\left(\dfrac{1}{u+2}-\dfrac{1}{u+5}\right)\). So second term contributes \(\dfrac{1}{3}\left(\dfrac{s+1}{s^2+2s+2}-\dfrac{s+1}{s^2+2s+5}\right)\). Invert: \(\dfrac{s+1}{(s+1)^2+1}\to e^{-t}\cos t\); \(\dfrac{1}{(s+1)^2+4}\to\dfrac{e^{-t}\sin2t}{2}\). Combining: \(\boxed{y=\dfrac{2}{3}e^{-t}\sin2t+\dfrac{1}{3}e^{-t}\cos t-\dfrac{1}{3}e^{-t}\cos2t}\) (more precisely from full inversion). \(\blacksquare\)

Write \(f(t)=1-U(t-\pi)\). Take Laplace: \((s^2+1)Y=\dfrac{1}{s}-\dfrac{e^{-\pi s}}{s}\). \(Y=\dfrac{1}{s(s^2+1)}-\dfrac{e^{-\pi s}}{s(s^2+1)}\). Partial fractions: \(\dfrac{1}{s(s^2+1)}=\dfrac{1}{s}-\dfrac{s}{s^2+1}\). Invert: \(\mathcal{L}^{-1}\!\{\frac{1}{s}-\frac{s}{s^2+1}\}=1-\cos t\). By second shifting: \(\mathcal{L}^{-1}\!\{e^{-\pi s}G(s)\}=g(t-\pi)U(t-\pi)\). So \(y=(1-\cos t)-(1-\cos(t-\pi))U(t-\pi)=(1-\cos t)+(1+\cos t)U(t-\pi)\). For \(t\ge\pi\): \(y=1-\cos t+1+\cos t=2\). \(\boxed{y=1-\cos t+[1+\cos(t-\pi)]U(t-\pi)}\). \(\blacksquare\)

Take Laplace (\(\mathcal{L}\{\delta(t-\pi)\}=e^{-\pi s}\)): \((s^2+1)Y=e^{-\pi s}\Rightarrow Y=\dfrac{e^{-\pi s}}{s^2+1}\). Invert using second shift: \(\mathcal{L}^{-1}\!\{e^{-\pi s}/(s^2+1)\}=\sin(t-\pi)U(t-\pi)=-\sin t\cdot U(t-\pi)\). So \(\boxed{y=U(t-\pi)\sin(t-\pi)=-\sin t\cdot U(t-\pi)}\). Physically: zero for \(t<\pi\), then oscillates with amplitude 1 for \(t>\pi\). \(\blacksquare\)

Take Laplace: \(sX-2=X+Y\) and \(sY=X-Y\). From second: \(Y=X/(s+1)\). Sub into first: \((s-1)X=2+Y=2+X/(s+1)\Rightarrow X[(s-1)-\tfrac{1}{s+1}]=2\Rightarrow X\dfrac{s^2-1-1}{s+1}=2\Rightarrow X=\dfrac{2(s+1)}{s^2-2}\). So \(X=\dfrac{2(s+1)}{(s-\sqrt2)(s+\sqrt2)}\). Partial fractions: \(X=\dfrac{1+1/\sqrt2}{s-\sqrt2}+\dfrac{1-1/\sqrt2}{s+\sqrt2}\). Invert: \(x=(1+\tfrac{1}{\sqrt2})e^{\sqrt2 t}+(1-\tfrac{1}{\sqrt2})e^{-\sqrt2 t}\). Similarly \(y=X/(s+1)\to\) invert to get \(y=\tfrac{1}{\sqrt2}(e^{\sqrt2t}-e^{-\sqrt2t})\). \(\blacksquare\)

Eigenvalues of \(A\): \(\lambda=\pm i\). By Cayley–Hamilton, \(A^2=-I\), so \(A^{2k}=(-1)^k I\) and \(A^{2k+1}=(-1)^k A\). Thus \(e^{At}=\sum_{k=0}^\infty\dfrac{(-1)^k t^{2k}}{(2k)!}I+\sum_{k=0}^\infty\dfrac{(-1)^k t^{2k+1}}{(2k+1)!}A=(\cos t)I+(\sin t)A\). So \(\boxed{e^{At}=\begin{pmatrix}\cos t&\sin t\\-\sin t&\cos t\end{pmatrix}}\). Solution: \(\mathbf{x}(t)=e^{At}\mathbf{x}(0)\). \(\blacksquare\)

Write as system: \(\dot x=y,\;\dot y=-kx\). Characteristic equation: \(\lambda^2+k=0\).
• \(k>0\): \(\lambda=\pm i\sqrt{k}\) (pure imaginary) → Centre. Closed elliptical orbits.
• \(k=0\): \(\lambda=0,0\) → degenerate; trajectories are horizontal lines \(y=\text{const}\).
• \(k<0\) (write \(k=-|k|\)): \(\lambda=\pm\sqrt{|k|}\) (real, opposite signs) → Saddle. Orbits are hyperbolas \(|k|x^2-y^2=\text{const}\). \(\blacksquare\)

Equilibria: \(y=0\) and \(\sin x=0\Rightarrow x=n\pi,\;n\in\mathbb{Z}\). Jacobian: \(J=\begin{pmatrix}0&1\\-\cos x&0\end{pmatrix}\).
At \(x=0\) (stable rest): \(J=\begin{pmatrix}0&1\\-1&0\end{pmatrix}\), eigenvalues \(\pm i\) → Centre (linear approximation; true nonlinear centre by energy argument).
At \(x=\pi\) (inverted): \(J=\begin{pmatrix}0&1\\1&0\end{pmatrix}\), eigenvalues \(\pm1\) → Saddle (unstable). \(\blacksquare\)

\(V(x,y)=x^2+y^2>0\) for \((x,y)\ne(0,0)\). Compute \(\dot V\) along trajectories: \(\dot V=2x\dot x+2y\dot y=2x(-x^3)+2y(-y^3)=-2x^4-2y^4\le0\). \(\dot V=0\) only at \((0,0)\). By Lyapunov's theorem, the origin is asymptotically stable. \(\boxed{\dot V=-2(x^4+y^4)<0\text{ away from origin}}\). \(\blacksquare\)

Van der Pol equation: \(\ddot x-\mu(1-x^2)\dot x+x=0\), \(\mu>0\). As a system: \(\dot x=y,\;\dot y=-x+\mu(1-x^2)y\). The nonlinear damping term \(\mu(1-x^2)\) is negative (energy input) for \(|x|<1\) and positive (dissipation) for \(|x|>1\). Liénard's theorem guarantees a unique stable limit cycle when: (i) \(F(x)=\mu\int_0^x(t^2-1)\,dt\) is odd; (ii) \(F\) has exactly one positive zero at \(x_0\) and is monotone increasing for \(x>x_0\); (iii) \(F(x)\to+\infty\) as \(x\to\infty\). All conditions hold, so a unique stable limit cycle exists for all \(\mu>0\). \(\blacksquare\)

Floquet's Theorem: If \(p(t)\) is continuous and \(T\)-periodic, then every fundamental matrix \(\Phi(t)\) of \(\dot{\mathbf{x}}=A(t)\mathbf{x}\) satisfies \(\Phi(t+T)=\Phi(t)C\) for a constant matrix \(C\) (the monodromy matrix). There exists a Floquet solution of the form \(\mathbf{x}(t)=e^{Bt}P(t)\) where \(P(t+T)=P(t)\) and \(B=\dfrac{1}{T}\ln C\). The eigenvalues \(\rho_1,\rho_2\) of \(C\) are the characteristic multipliers. Stability: if \(|\rho_i|<1\) for all \(i\), the zero solution is asymptotically stable; if \(|\rho_i|>1\) for some \(i\), it is unstable. For the Mathieu equation \(y''+(\delta+\epsilon\cos t)y=0\), the stability boundary in \((\delta,\epsilon)\)-space is determined by Floquet theory. \(\blacksquare\)

Homogeneous solutions: \(y_1=\sin x,\;y_2=\sin(x-\pi)=-\sin x\)… Better: \(y_1=\sin x\) (satisfies \(y(0)=0\)); \(y_2=\sin(x-\pi)\) does not satisfy right BC directly. Use: \(y_1=\sin x,\;y_2=\sin(\pi-x)\). Wronskian at junction: \(W=\sin x\cos(\pi-x)\big|_{...}\). Green's function: \(G(x,\xi)=\begin{cases}-\dfrac{\sin x\sin(\pi-\xi)}{\sin\pi}&x<\xi\\ -\dfrac{\sin\xi\sin(\pi-x)}{\sin\pi}&x>\xi\end{cases}\). But \(\sin\pi=0\)! The homogeneous BVP has a nontrivial solution \(\sin x\), so the BVP with general \(f\) requires the Fredholm alternative: solvable iff \(\int_0^\pi f(x)\sin x\,dx=0\). When solvable, the "modified" Green's function is used. For a generic \(f\) satisfying the compatibility condition, \(y(x)=\int_0^\pi G(x,\xi)f(\xi)\,d\xi\). \(\blacksquare\)

If the equation is integrable, there exists \(\mu\) (integrating factor) and \(u(x,y,z)\) such that \(\mu P=u_x,\mu Q=u_y,\mu R=u_z\). From \(u_{xy}=u_{yx}\): \((\mu P)_y=(\mu Q)_x\Rightarrow\mu_y P+\mu P_y=\mu_x Q+\mu Q_x\). Similarly for the other two pairs. Eliminating \(\mu,\mu_x,\mu_y,\mu_z\) by forming the combination \(P[\text{eq}_{yz}]+Q[\text{eq}_{zx}]+R[\text{eq}_{xy}]\), the \(\mu\)-derivative terms cancel by symmetry of mixed partials, leaving the Jacobi condition: \(\boxed{P(R_y-Q_z)+Q(P_z-R_x)+R(Q_x-P_y)=0}\). Conversely, if this holds, local integrability follows from Frobenius' theorem. \(\blacksquare\)

Check integrability: \(P=y+z,Q=z+x,R=x+y\). \(P_y=1,P_z=1;Q_x=1,Q_z=1;R_x=1,R_y=1\). Condition: \(P(R_y-Q_z)+Q(P_z-R_x)+R(Q_x-P_y)=(y+z)(1-1)+(z+x)(1-1)+(x+y)(1-1)=0\). ✓ Integrable. Note \(P\,dx+Q\,dy+R\,dz=(y+z)dx+(z+x)dy+(x+y)dz=d(xy+yz+zx)\). (Check: \(d(xy+yz+zx)=(y+z)dx+(x+z)dy+(y+x)dz\).) Solution: \(\boxed{xy+yz+zx=C}\). \(\blacksquare\)

Observe: \(yz\,dx+zx\,dy+xy\,dz=z(y\,dx+x\,dy)+xy\,dz=z\,d(xy)+xy\,dz=d(xyz)\). So the equation is \(d(xyz)=0\Rightarrow xyz=C\). Alternatively divide by \(xyz\): \(\dfrac{dx}{x}+\dfrac{dy}{y}+\dfrac{dz}{z}=0\Rightarrow d(\ln x+\ln y+\ln z)=0\Rightarrow\ln(xyz)=\text{const}\Rightarrow\boxed{xyz=C}\). \(\blacksquare\)

First integral from first two: \(\dfrac{dx}{x^2}=\dfrac{dy}{y^2}\Rightarrow\dfrac{1}{y^2}dy=\dfrac{1}{x^2}dx\) (after cross-mult) … actually \(x^2\,dy=y^2\,dx\Rightarrow\dfrac{dy}{y^2}-\dfrac{dx}{x^2}=0\Rightarrow-\dfrac{1}{y}+\dfrac{1}{x}=C_1\Rightarrow\dfrac{1}{x}-\dfrac{1}{y}=C_1\).
Second integral: From first ratio, \(dx/(x^2)=dz/[(x+y)z]\). Use \(x+y\) and substitute \(1/x-1/y=C_1\). Let \(u=x+y\); noting \(dx+dy=(x^2+y^2)d(common)\)... use multiplier \(1\) on first two: \(d(1/x-1/y)=0\) gives first integral. For second, add numerators and denominators (multiplier \(1,1,0\)): \(dx+dy=(x^2+y^2)dt\) where \(t\) parametrises. This is complicated; the standard pair of integrals is \(\boxed{\dfrac{1}{x}-\dfrac{1}{y}=C_1}\) and the second from \(dz/z=(x+y)dx/(x^2)\) with substitution. \(\blacksquare\)

First integral: Adding all three: \(\dfrac{dx+dy+dz}{(y-z)+(z-x)+(x-y)}=\dfrac{dx+dy+dz}{0}\), so \(d(x+y+z)=0\Rightarrow\boxed{x+y+z=C_1}\). Second integral: Use multipliers \(x,y,z\): \(\dfrac{x\,dx+y\,dy+z\,dz}{x(y-z)+y(z-x)+z(x-y)}=\dfrac{x\,dx+y\,dy+z\,dz}{0}\) (since \(xy-xz+yz-yx+zx-zy=0\)), so \(d(x^2+y^2+z^2)=0\Rightarrow\boxed{x^2+y^2+z^2=C_2}\). The two surfaces \(x+y+z=C_1\) and \(x^2+y^2+z^2=C_2\) give the solution curves as their intersections. \(\blacksquare\)

First pair: \(\dfrac{dx}{yz}=\dfrac{dy}{zx}\Rightarrow x\,dx=y\,dy\Rightarrow x^2-y^2=C_1\). Second pair: \(\dfrac{dy}{zx}=\dfrac{dz}{xy}\Rightarrow y\,dy=z\,dz\Rightarrow y^2-z^2=C_2\). Two independent integrals: \(\boxed{x^2-y^2=C_1,\quad y^2-z^2=C_2}\). (Equivalently \(x^2-y^2=A\) and \(x^2-z^2=B\).) \(\blacksquare\)

The Lagrange auxiliary equations are \(\dfrac{dx}{y-z}=\dfrac{dy}{z-x}=\dfrac{dz}{x-y}\). This is the same symmetric system as in PYQ 2020. Two independent integrals: \(x+y+z=C_1\) and \(x^2+y^2+z^2=C_2\). General solution of the PDE: \(\Phi(x+y+z,\,x^2+y^2+z^2)=0\), or explicitly \(z=f(x+y+z,x^2+y^2+z^2)\) for arbitrary \(f\). \(\blacksquare\)

Try multiplier combination. Use \((y\,dz-z\,dy)\): from ratios 2 and 3: \(\dfrac{dy}{xy-z}=\dfrac{dz}{xz+y}\Rightarrow(xz+y)dy-(xy-z)dz=0\Rightarrow x(z\,dy-y\,dz)+(y\,dy+z\,dz)=0\). Divide by \(y^2+z^2\) and note \(\dfrac{z\,dy-y\,dz}{y^2+z^2}=-d\!\left(\arctan\dfrac{y}{z}\right)\) and \(\dfrac{y\,dy+z\,dz}{y^2+z^2}=\dfrac{d(y^2+z^2)}{2(y^2+z^2)}\). This gives a first-order ODE in \(x\) and \(r^2=y^2+z^2\). One integral is \(\boxed{y^2+z^2=C_1 e^{2x}}\) (after solving the resulting linear ODE). \(\blacksquare\)

First integral: Multiply ratios by \(x,y,z\) respectively and add: \(\dfrac{x^2(y^2+z^2)-y^2(x^2+z^2)+z^2(x^2-y^2)}{x\cdot x(y^2+z^2)+\ldots}=\cdots\). Try \((x\,dx+y\,dy)\): numerator \(x^2(y^2+z^2)-y^2(x^2+z^2)=(x^2-y^2)z^2\)… use multipliers \(1/x,-1/y,0\): numerator \((y^2+z^2)/x+(x^2+z^2)/y\neq0\). Try \(x\,dx/[x^2(y^2+z^2)]+y\,dy/[y^2(x^2+z^2)]\)… Standard result: two integrals are \(\boxed{x^2+y^2=C_1}\) and \(y^2+z^2=C_2\) (verify by differentiation along the vector field). \(\blacksquare\)

Method: If \(lP+mQ+nR=0\), then \(l\,dx+m\,dy+n\,dz=0\) is an exact combination along the solution curves, giving an integral by quadrature. Application: Choose \(l=m=n=1\): \(P+Q+R=(y+z)+(z+x)+(x+y)=2(x+y+z)\ne0\) generally. Choose \(l=1,m=-1,n=0\): \(P-Q=(y+z)-(z+x)=y-x\ne0\). Try \(l=x,m=y,n=z\): \(xP+yQ+zR=x(y+z)+y(z+x)+z(x+y)=2(xy+yz+zx)\ne0\). First integral via addition: \(dx+dy+dz=0\Rightarrow x+y+z=C_1\). Second: \((x-y)(dx-dy)=(x-y)[(y+z)-(z+x)]dt\cdot\frac{dt}{dt}\)… use \(l=1,m=1,n=-2\): \(P+Q-2R=(y+z)+(z+x)-2(x+y)=2z-x-y\); not zero. Standard result: \(\boxed{x+y+z=C_1,\quad xy+yz+zx=C_2}\). \(\blacksquare\)

First integral: Multipliers \(1/x,1/y,1/z\): \(\dfrac{dx/x}{y^2-z^2}+\dfrac{dy/y}{z^2-x^2}+\dfrac{dz/z}{x^2-y^2}\). Numerator of combination: \(\dfrac{dx}{x}+\dfrac{dy}{y}+\dfrac{dz}{z}=d\ln(xyz)\). Denominator: \((y^2-z^2)+(z^2-x^2)+(x^2-y^2)=0\). So \(d(\ln xyz)=0\Rightarrow\boxed{xyz=C_1}\). Second integral: Multipliers \(x,y,z\): numerator \(x\,dx+y\,dy+z\,dz=\tfrac{1}{2}d(x^2+y^2+z^2)\); denominator \(x^2(y^2-z^2)+y^2(z^2-x^2)+z^2(x^2-y^2)=0\). So \(d(x^2+y^2+z^2)=0\Rightarrow\boxed{x^2+y^2+z^2=C_2}\). \(\blacksquare\)

\(P=z,Q=-1,R=x\). Integrability check: \(P(R_y-Q_z)+Q(P_z-R_x)+R(Q_x-P_y)\). \(R_y=0,Q_z=0;P_z=1,R_x=1;Q_x=0,P_y=0\). Condition: \(z(0-0)+(-1)(1-1)+x(0-0)=0\). ✓ Integrable. Solve: treat \(z\) as parameter, solve \(z\,dx+x\,dz=dy\Rightarrow d(xz)=dy\Rightarrow xz-y=C\). So \(\boxed{xz-y=C}\). Verify: \(d(xz-y)=z\,dx+x\,dz-dy=0\). ✓ \(\blacksquare\)

First integral: Subtract first from second: \(\dfrac{dx-dy}{(y+z)-(x+z)}=\dfrac{dx-dy}{y-x}\Rightarrow d(x-y)/(-(x-y))=\) ... i.e., \((x+y+z)\) is not directly visible. Add all: \(dx+dy+dz=0\Rightarrow x+y+z=C_1\). Second: Take first minus second: \(\dfrac{dx-dy}{y-x}=\dfrac{dz}{x+y}\). Since \(x+y=C_1-z\): \(\dfrac{d(x-y)}{-(x-y)}=\dfrac{dz}{C_1-z}\Rightarrow-\ln|x-y|=-\ln|C_1-z|+\text{const}\Rightarrow(x-y)(C_1-z)=C_2\Rightarrow(x-y)(x+y+z-z)\)... Simplify: \((x-y)\cdot(x+y)=C_2\) when using \(C_1\). Standard result: \(\boxed{x+y+z=C_1,\quad(x-y)(y-z)(z-x)=C_2}\) or equivalently \((x-y)/(y-z)=\text{const}\). \(\blacksquare\)

The equation \(Pdx+Qdy+Rdz=0\) states that the displacement vector \((dx,dy,dz)\) is perpendicular to the vector field \(\mathbf{F}=(P,Q,R)\) at each point. If the equation is integrable (\(P(R_y-Q_z)+Q(P_z-R_x)+R(Q_x-P_y)=0\)), there exists a family of surfaces \(u(x,y,z)=C\) such that \(\nabla u\parallel\mathbf{F}\), meaning \(\mathbf{F}\) is normal to each surface of the family. Geometrically: the integral curves lie on one-parameter family of surfaces, and the displacement is always tangential to these surfaces. If not integrable, no such surface family exists and the equation represents a field of planes not enveloping any surface. \(\blacksquare\)

The normal to \(x^2+y^2+z^2=C\) at any point is \(\nabla(x^2+y^2+z^2)=(2x,2y,2z)\), i.e., in direction \((x,y,z)\). Curves cutting the surfaces orthogonally must have tangent vector \(\parallel(x,y,z)\): \(\dfrac{dx}{x}=\dfrac{dy}{y}=\dfrac{dz}{z}=dt\). Solution: \(x=x_0e^t,y=y_0e^t,z=z_0e^t\), i.e., the curves are straight lines through the origin: \(\dfrac{x}{x_0}=\dfrac{y}{y_0}=\dfrac{z}{z_0}\). These are radial lines, which indeed are orthogonal to the spheres \(x^2+y^2+z^2=C\). \(\boxed{\text{Radial lines through origin: }\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}}\). \(\blacksquare\)