← 2022 Paper 1

UPSC 2022 Maths Optional Paper 1 Q5a — Step-by-Step Solution

10 marks · Section B

Linear first-order · ODEs · asked 6× in 14 yrs · Read the full method →

Question

Show that the general solution of dydx+Py=Q\dfrac{dy}{dx}+Py=Q can be written as

y=QP−e−∫P dx ⁣{C+∫e∫P dx d ⁣(QP)},y=\dfrac{Q}{P}-e^{-\int P\,dx}\!\left\{C+\int e^{\int P\,dx}\,d\!\left(\dfrac{Q}{P}\right)\right\},

where P,QP,Q are non-zero functions of xx and CC is an arbitrary constant.

Technique

Standard integrating-factor solution; integration by parts rearranges the integral to extract Q/PQ/P as an “equilibrium” term plus the homogeneous solution.

Solution

Setup. The standard solution of y′+Py=Qy'+Py=Q uses integrating factor μ=e∫P dx\mu=e^{\int P\,dx}:

(μy)′=μQ,μy=∫μQ dx+C0,y=e−∫P dx ⁣[∫e∫P dxQ dx+C0].(\mu y)'=\mu Q,\quad\mu y=\int\mu Q\,dx+C_0,\quad y=e^{-\int P\,dx}\!\left[\int e^{\int P\,dx}Q\,dx+C_0\right].

We need to rewrite this as the given form.

Step 1 — Integrate by parts on ∫μQ dx\int\mu Q\,dx

In ∫e∫P dx⋅Q dx\int e^{\int P\,dx}\cdot Q\,dx, write Q=P⋅(Q/P)Q=P\cdot(Q/P). Then

∫e∫P dx⋅P⋅QP dx.\int e^{\int P\,dx}\cdot P\cdot\dfrac{Q}{P}\,dx.

Recognise e∫P dxP dx=d(e∫P dx)e^{\int P\,dx}P\,dx=d(e^{\int P\,dx}) since ddxe∫P dx=Pe∫P dx\dfrac{d}{dx}e^{\int P\,dx}=P e^{\int P\,dx}.

So ∫e∫P dxP⋅QP dx=∫QP d(e∫P dx)\int e^{\int P\,dx}P\cdot\dfrac{Q}{P}\,dx=\int\dfrac{Q}{P}\,d(e^{\int P\,dx}).

Step 2 — Apply integration by parts

∫u dv=uv−∫v du\int u\,dv=uv-\int v\,du with u=Q/Pu=Q/P, dv=d(e∫P dx)dv=d(e^{\int P\,dx}):

∫QP d(e∫P dx)=QP⋅e∫P dx−∫e∫P dx d ⁣(QP).\int\dfrac{Q}{P}\,d(e^{\int P\,dx})=\dfrac{Q}{P}\cdot e^{\int P\,dx}-\int e^{\int P\,dx}\,d\!\left(\dfrac{Q}{P}\right).

Step 3 — Assemble

μy=∫μQ dx+C0=QPμ−∫μ d(Q/P)+C0\mu y=\int\mu Q\,dx+C_0=\dfrac{Q}{P}\mu-\int\mu\,d(Q/P)+C_0.

Divide by μ=e∫P dx\mu=e^{\int P\,dx}:

y=QP−e−∫P dx ⁣[∫e∫P dx d(Q/P)−C0].y=\dfrac{Q}{P}-e^{-\int P\,dx}\!\left[\int e^{\int P\,dx}\,d(Q/P)-C_0\right].

Set C=−C0C=-C_0 (a different arbitrary constant):

y=QP−e−∫P dx ⁣[∫e∫P dx d(Q/P)+C]y=\dfrac{Q}{P}-e^{-\int P\,dx}\!\left[\int e^{\int P\,dx}\,d(Q/P)+C\right] =QP−e−∫P dx ⁣{C+∫e∫P dx d ⁣(QP)}.■=\dfrac{Q}{P}-e^{-\int P\,dx}\!\left\{C+\int e^{\int P\,dx}\,d\!\left(\dfrac{Q}{P}\right)\right\}.\qquad\blacksquare

Answer

  y=QP−e−∫P dx ⁣{C+∫e∫P dx d(Q/P)}.  \boxed{\;y=\dfrac{Q}{P}-e^{-\int P\,dx}\!\left\{C+\int e^{\int P\,dx}\,d(Q/P)\right\}.\;}
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