← 2022 Paper 1

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

10 marks · Section B

Linear first-order · ODEs · asked 5× in 13 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=QPePdx ⁣{C+ePdxd ⁣(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=ePdx\mu=e^{\int P\,dx}:

(μy)=μQ,μy=μQdx+C0,y=ePdx ⁣[ePdxQdx+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 μQdx\int\mu Q\,dx

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

ePdxPQPdx.\int e^{\int P\,dx}\cdot P\cdot\dfrac{Q}{P}\,dx.

Recognise ePdxPdx=d(ePdx)e^{\int P\,dx}P\,dx=d(e^{\int P\,dx}) since ddxePdx=PePdx\dfrac{d}{dx}e^{\int P\,dx}=P e^{\int P\,dx}.

So ePdxPQPdx=QPd(ePdx)\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

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

QPd(ePdx)=QPePdxePdxd ⁣(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=μQdx+C0=QPμμd(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=ePdx\mu=e^{\int P\,dx}:

y=QPePdx ⁣[ePdxd(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=QPePdx ⁣[ePdxd(Q/P)+C]y=\dfrac{Q}{P}-e^{-\int P\,dx}\!\left[\int e^{\int P\,dx}\,d(Q/P)+C\right] =QPePdx ⁣{C+ePdxd ⁣(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=QPePdx ⁣{C+ePdxd(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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