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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 dxdy+Py=Q can be written as
y=PQ−e−∫Pdx{C+∫e∫Pdxd(PQ)},
where P,Q are non-zero functions of x and C is an arbitrary constant.
Technique
Standard integrating-factor solution; integration by parts rearranges the integral to extract Q/P as an “equilibrium” term plus the homogeneous solution.
Solution
Setup. The standard solution of y′+Py=Q uses integrating factor μ=e∫Pdx:
(μy)′=μQ,μy=∫μQdx+C0,y=e−∫Pdx[∫e∫PdxQdx+C0].
We need to rewrite this as the given form.
Step 1 — Integrate by parts on ∫μQdx
In ∫e∫Pdx⋅Qdx, write Q=P⋅(Q/P). Then
∫e∫Pdx⋅P⋅PQdx.
Recognise e∫PdxPdx=d(e∫Pdx) since dxde∫Pdx=Pe∫Pdx.
So ∫e∫PdxP⋅PQdx=∫PQd(e∫Pdx).
Step 2 — Apply integration by parts
∫udv=uv−∫vdu with u=Q/P, dv=d(e∫Pdx):
∫PQd(e∫Pdx)=PQ⋅e∫Pdx−∫e∫Pdxd(PQ).
Step 3 — Assemble
μy=∫μQdx+C0=PQμ−∫μd(Q/P)+C0.
Divide by μ=e∫Pdx:
y=PQ−e−∫Pdx[∫e∫Pdxd(Q/P)−C0].
Set C=−C0 (a different arbitrary constant):
y=PQ−e−∫Pdx[∫e∫Pdxd(Q/P)+C]
=PQ−e−∫Pdx{C+∫e∫Pdxd(PQ)}.■
Answer
y=PQ−e−∫Pdx{C+∫e∫Pdxd(Q/P)}.