← 2014 Paper 1
UPSC 2014 Maths Optional Paper 1 Q6a — Step-by-Step Solution
10 marks · Section B
Method of variation of parameters · ODEs · asked 11× in 13 yrs · Read the full method →
Question
Solve by the method of variation of parameters:
dxdy−5y=sinx.
Technique
Variation of parameters for first-order linear ODE; standard ∫eaxsinbx formula.
Solution
Strategy. For first-order linear ODE y′+P(x)y=Q(x), variation of parameters replaces the integration constant C in the homogeneous solution yh=Ce−∫P by an unknown function u(x).
Step 1 — Homogeneous solution
yh′=5yh⇒yh=Ce5x. So a fundamental solution is y1=e5x.
Step 2 — Variation: try yp=u(x)e5x
yp′=u′e5x+5ue5x. Substitute into y′−5y=sinx:
u′e5x+5ue5x−5ue5x=sinx⟹u′e5x=sinx⟹u′=e−5xsinx.
Step 3 — Integrate
Use the standard formula
∫eaxsin(bx)dx=a2+b2eax(asinbx−bcosbx).
With a=−5,b=1:
u=∫e−5xsinxdx=25+1e−5x(−5sinx−cosx)=−26e−5x(5sinx+cosx).
(Plus a constant of integration absorbed into the general solution.)
Step 4 — Particular and general solution
yp=u⋅e5x=−265sinx+cosx.
General solution:
Answer
y=Ce5x−265sinx+cosx.