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CBSE Class 12 · 2009 · Differential EquationsLong Answer

Solve the following differential equation:

(1 + x^{2}) \frac{d y}{d x} + y = tan^{- 1} x

Answer: $y = tan^{- 1} x - 1 + C e^{- t a n^{- 1} x}$

Solution

Divide by

(1 + x^{2})

:

\frac{d y}{d x} + \frac{1}{1 + x ^{2}} y = \frac{tan ^{- 1} x}{1 + x ^{2}}

I.F. = e^{\int \frac{d x}{1 + x ^{2}}} = e^{t a n^{- 1} x}

y e^{t a n^{- 1} x} = \int e^{t a n^{- 1} x} \frac{tan ^{- 1} x}{1 + x ^{2}} d x

Put

t = tan^{- 1} x

:

\int t e^{t} d t = e^{t} (t - 1)

, so

y e^{t a n^{- 1} x} = e^{t a n^{- 1} x} (tan^{- 1} x - 1) + C \Rightarrow y = tan^{- 1} x - 1 + C e^{- t a n^{- 1} x}

CBSE Class 12 Board · 2009 · Subjective

Practise more: all CBSE board PYQs on Differential Equations →

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