Ordinary Differential Equations

UPSC Mathematics Optional · Paper I · All Methods & Theorems

1. First Order — Separable, Exact, Linear 2. Bernoulli, Clairaut, Singular Solutions 3. Second Order — CF & PI 4. Variation of Parameters 5. Cauchy–Euler Equation 6. Systems of ODEs 7. Series Solutions — Frobenius 8. Bessel's & Legendre's Equations 9. Existence & Uniqueness

§1 · First Order ODEs

Separable Equations

\(f(y)\,dy = g(x)\,dx\). Integrate both sides. Check constant solution \(f(y)=0\).

Exact Equations PYQ★★★

\(M\,dx + N\,dy = 0\) is exact iff \(\dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x}\).

Solution: \(F(x,y)=C\) where \(F_x=M,\; F_y=N\).

Find: \(F = \int M\,dx + \phi(y)\), then use \(F_y=N\) to find \(\phi'(y)\).

Integrating Factor (IF) PYQ★★★

If not exact, multiply by IF \(\mu\) to make it exact.

If \(\dfrac{M_y-N_x}{N}=f(x)\) only, then \(\mu=e^{\int f(x)dx}\).
If \(\dfrac{N_x-M_y}{M}=g(y)\) only, then \(\mu=e^{\int g(y)dy}\).
If \(Mx+Ny=0\): \(\mu=\dfrac{1}{Mx-Ny}\). If \(Mx+Ny=k\): \(\mu=\dfrac{1}{Mx+Ny}\).

Linear First Order: \(\dfrac{dy}{dx}+P(x)y=Q(x)\) PYQ★★★

IF \(= e^{\int P\,dx}\). Solution: \(y\cdot e^{\int P\,dx} = \int Q\,e^{\int P\,dx}\,dx + C\).

Homogeneous Equations

\(\dfrac{dy}{dx}=f\!\left(\dfrac{y}{x}\right)\). Substitute \(v=y/x\) → separable in \(v\) and \(x\).

§2 · Bernoulli, Clairaut & Singular Solutions

Bernoulli Equation PYQ★★

\(\dfrac{dy}{dx}+P(x)y=Q(x)y^n\). Substitute \(v=y^{1-n}\) → linear in \(v\).

\(\dfrac{dv}{dx}+(1-n)Pv=(1-n)Q\).

Clairaut's Equation PYQ★★★

\(y = xp + f(p)\) where \(p=dy/dx\). General solution: replace \(p\) by \(c\): \(y=cx+f(c)\) (family of lines).

Singular solution: differentiate w.r.t. \(x\) and eliminate \(p\): \(x+f'(p)=0\), then use \(y=xp+f(p)\).

Geometrically: singular solution = envelope of the family of lines.

p-discriminant & c-discriminant (Singular Solutions) PYQ★★

Eliminate \(p\) between \(F(x,y,p)=0\) and \(\partial F/\partial p=0\) → p-discriminant relation.

Eliminate \(c\) between \(F(x,y,c)=0\) and \(\partial F/\partial c=0\) → c-discriminant relation.

Singular solution satisfies both. TAC locus and node locus also appear — check which is genuine.

§3 · Second Order Linear ODEs with Constant Coefficients

Complementary Function (CF) PYQ★★★

For \((D^2+aD+b)y=0\), auxiliary equation: \(m^2+am+b=0\).

Distinct real roots \(m_1,m_2\): CF \(=c_1e^{m_1x}+c_2e^{m_2x}\).
Equal roots \(m_1=m_2=m\): CF \(=(c_1+c_2x)e^{mx}\).
Complex roots \(\alpha\pm i\beta\): CF \(=e^{\alpha x}(c_1\cos\beta x+c_2\sin\beta x)\).

For \(n\)-th order: analogous with \(n\) roots, repeated roots give poly factor \((c_1+c_2x+\cdots+c_kx^{k-1})\).

Particular Integral (PI) — Operator Method PYQ★★★

\(\text{PI} = \dfrac{1}{f(D)}X\). Key cases:

\(\dfrac{1}{f(D)}e^{ax} = \dfrac{e^{ax}}{f(a)}\), provided \(f(a)\ne 0\).
If \(f(a)=0\): \(\dfrac{1}{(D-a)^r}e^{ax} = \dfrac{x^r e^{ax}}{r!}\).
\(\dfrac{1}{f(D)}\sin(ax+b) = \dfrac{\sin(ax+b)}{f(-a^2)}\), replace \(D^2\to -a^2\), if \(f(-a^2)\ne 0\).
If denominator = 0: multiply numerator by \(x\) and differentiate denominator (use exponential shift).
\(\dfrac{1}{f(D)}x^m\): expand \([f(D)]^{-1}\) in ascending powers of \(D\) and operate.
\(\dfrac{1}{f(D)}e^{ax}V = e^{ax}\dfrac{1}{f(D+a)}V\) (exponential shift).
\(\dfrac{1}{f(D)}xV = x\dfrac{1}{f(D)}V - \dfrac{f'(D)}{[f(D)]^2}V\).

§4 · Variation of Parameters

Variation of Parameters PYQ★★★

For \(y''+P(x)y'+Q(x)y=R(x)\) with CF \(y=c_1y_1+c_2y_2\):

\[\text{PI} = -y_1\int\frac{y_2 R}{W}\,dx + y_2\int\frac{y_1 R}{W}\,dx\]

where \(W = \begin{vmatrix}y_1&y_2\\y_1'&y_2'\end{vmatrix} = y_1y_2'-y_2y_1'\) (Wronskian, \(\ne 0\)).

Wronskian — Linear Independence PYQ★★

\(W(y_1,y_2)\ne 0\) on interval ⟹ \(y_1,y_2\) linearly independent solutions.

Abel's identity: \(W(x)=W(x_0)\exp\!\left(-\int_{x_0}^x P(t)\,dt\right)\). So \(W\) is either identically 0 or never 0.

Reduction of Order — One Known Solution

If \(y_1\) is one solution of \(y''+Py'+Qy=0\), put \(y=vy_1\) → separable ODE for \(v'\).

\(v' = \dfrac{1}{y_1^2}\exp\!\left(-\int P\,dx\right)\) up to constant.

§5 · Cauchy–Euler (Equidimensional) Equation

Cauchy–Euler Equation PYQ★★★

\(x^n y^{(n)}+a_1 x^{n-1}y^{(n-1)}+\cdots+a_n y = f(x)\).

Substitution: \(x=e^t\) (i.e. \(t=\ln x\)) transforms to constant-coefficient ODE.

\(x\dfrac{d}{dx}=D_t\), \(x^2\dfrac{d^2}{dx^2}=D_t(D_t-1)\), \(x^3\dfrac{d^3}{dx^3}=D_t(D_t-1)(D_t-2)\).

Trial \(y=x^m\): substituting gives indicial equation in \(m\). Same cases as constant coefficient.

§6 · Systems of Linear ODEs

Operator Method for Simultaneous ODEs PYQ★★★

For \(\dfrac{dx}{dt}=ax+by\), \(\dfrac{dy}{dt}=cx+dy\): operate with \(D=d/dt\).

Eliminate one variable (e.g. eliminate \(y\)) → single ODE in \(x\). Solve, then use first equation to find \(y\).

Note: the arbitrary constants are related — don't introduce extra ones.

Matrix Method: \(\mathbf{X}'=A\mathbf{X}\)

Solution: \(\mathbf{X} = e^{At}\mathbf{X}_0\). Compute via eigendecomposition.

If \(A\) diagonalizable: \(e^{At}=Pe^{\Lambda t}P^{-1}\) where \(\Lambda=\text{diag}(\lambda_i)\).

§7 · Series Solutions & Frobenius Method

Ordinary & Singular Points

\(y''+P(x)y'+Q(x)y=0\). \(x=a\) is ordinary if \(P,Q\) analytic there; singular if not.

Regular singular point: \((x-a)P(x)\) and \((x-a)^2Q(x)\) are both analytic at \(x=a\).

Frobenius Method PYQ★★★

At regular singular point \(x=0\), assume \(y=\sum_{n=0}^\infty a_n x^{n+r}\), \(a_0\ne 0\).

Substituting gives indicial equation in \(r\). Let roots be \(r_1\ge r_2\).

\(r_1-r_2\notin\mathbb{Z}\): two independent Frobenius series.
\(r_1=r_2\): second solution \(y_2=y_1\ln x + \sum b_n x^{n+r_1}\).
\(r_1-r_2\in\mathbb{Z}^+\): second solution may involve \(\ln x\).

§8 · Bessel's & Legendre's Equations

Bessel's Equation PYQ★★★

\(x^2y''+xy'+(x^2-n^2)y=0\). Solutions: Bessel functions \(J_n(x)\) and \(Y_n(x)\).

\[J_n(x)=\sum_{m=0}^\infty\frac{(-1)^m}{m!\,\Gamma(m+n+1)}\left(\frac{x}{2}\right)^{2m+n}\]

Recurrence: \(J_{n-1}(x)+J_{n+1}(x)=\dfrac{2n}{x}J_n(x)\).

\(J_{n-1}(x)-J_{n+1}(x)=2J_n'(x)\).

\(\dfrac{d}{dx}[x^n J_n]=x^n J_{n-1}\). \(\dfrac{d}{dx}[x^{-n}J_n]=-x^{-n}J_{n+1}\).

Special: \(J_{1/2}(x)=\sqrt{\dfrac{2}{\pi x}}\sin x\). \(J_{-1/2}(x)=\sqrt{\dfrac{2}{\pi x}}\cos x\).

Legendre's Equation PYQ★★★

\((1-x^2)y''-2xy'+n(n+1)y=0\). Solutions: \(P_n(x)\) (Legendre polynomial), \(Q_n(x)\).

\(P_n(x)=\dfrac{1}{2^n n!}\dfrac{d^n}{dx^n}(x^2-1)^n\) (Rodrigues' formula).

Values: \(P_0=1,\;P_1=x,\;P_2=\tfrac{1}{2}(3x^2-1),\;P_3=\tfrac{1}{2}(5x^3-3x)\).

Orthogonality: \(\int_{-1}^{1}P_m(x)P_n(x)\,dx=\begin{cases}0 & m\ne n\\\dfrac{2}{2n+1} & m=n\end{cases}\)

Recurrence: \((n+1)P_{n+1}=(2n+1)xP_n-nP_{n-1}\).

\(P_n(1)=1\). \(P_n(-1)=(-1)^n\). \(P_n(-x)=(-1)^nP_n(x)\).

§9 · Existence & Uniqueness

Picard's Existence–Uniqueness Theorem PYQ★★

For \(y'=f(x,y)\), \(y(x_0)=y_0\): if \(f\) is continuous and satisfies a Lipschitz condition in \(y\) on some rectangle \(|x-x_0|\le a\), \(|y-y_0|\le b\), then a unique solution exists on some interval about \(x_0\).

Lipschitz condition: \(|f(x,y_1)-f(x,y_2)|\le K|y_1-y_2|\).

Picard's Iteration Method

\(y_{n+1}(x) = y_0 + \int_{x_0}^x f(t,y_n(t))\,dt\). Successive approximations converge to unique solution.