← 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 12× in 14 yrs · Read the full method →

Question

Solve by the method of variation of parameters:

dydx−5y=sin⁡x.\dfrac{dy}{dx}-5y=\sin x.

Technique

Variation of parameters for first-order linear ODE; standard ∫eaxsin⁡bx\int e^{ax}\sin bx formula.

Solution

Strategy. For first-order linear ODE y′+P(x)y=Q(x)y'+P(x)y=Q(x), variation of parameters replaces the integration constant CC in the homogeneous solution yh=Ce−∫Py_h=Ce^{-\int P} by an unknown function u(x)u(x).

Step 1 — Homogeneous solution

yh′=5yh⇒yh=Ce5xy_h'=5y_h\Rightarrow y_h=Ce^{5x}. So a fundamental solution is y1=e5xy_1=e^{5x}.

Step 2 — Variation: try yp=u(x)e5xy_p=u(x)e^{5x}

yp′=u′e5x+5ue5xy_p'=u'e^{5x}+5ue^{5x}. Substitute into y′−5y=sin⁡xy'-5y=\sin x:

u′e5x+5ue5x−5ue5x=sin⁡x  ⟹  u′e5x=sin⁡x  ⟹  u′=e−5xsin⁡x.u'e^{5x}+5ue^{5x}-5ue^{5x}=\sin x\;\Longrightarrow\;u'e^{5x}=\sin x\;\Longrightarrow\;u'=e^{-5x}\sin x.

Step 3 — Integrate

Use the standard formula

∫eaxsin⁡(bx) dx=eax(asin⁡bx−bcos⁡bx)a2+b2.\int e^{ax}\sin(bx)\,dx=\dfrac{e^{ax}(a\sin bx-b\cos bx)}{a^{2}+b^{2}}.

With a=−5, b=1a=-5,\,b=1:

u=∫e−5xsin⁡x dx=e−5x(−5sin⁡x−cos⁡x)25+1=−e−5x(5sin⁡x+cos⁡x)26.u=\int e^{-5x}\sin x\,dx=\dfrac{e^{-5x}(-5\sin x-\cos x)}{25+1}=-\dfrac{e^{-5x}(5\sin x+\cos x)}{26}.

(Plus a constant of integration absorbed into the general solution.)

Step 4 — Particular and general solution

yp=u⋅e5x=−5sin⁡x+cos⁡x26.y_p=u\cdot e^{5x}=-\dfrac{5\sin x+\cos x}{26}.

General solution:

Answer

  y=Ce5x−5sin⁡x+cos⁡x26.  \boxed{\;y=Ce^{5x}-\dfrac{5\sin x+\cos x}{26}.\;}
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