Solve the differential equation $\left(1+x^{2}\right)\dfrac{dy}{dx}+y=e^{\tan^{-1}x}$.

$y\,e^{\tan^{-1}x}=\dfrac{1}{2}e^{2\tan^{-1}x}+C$ (i.e. $y=\dfrac{1}{2}e^{\tan^{-1}x}+Ce^{-\tan^{-1}x}$)

CBSE Class 12 · 2014 · Differential EquationsLong Answer

Solve the differential equation

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

Answer: $y e^{t a n^{- 1} x} = \frac{1}{2} e^{2 t a n^{- 1} x} + C$ (i.e. $y = \frac{1}{2} e^{t a n^{- 1} x} + C e^{- t a n^{- 1} x}$ )

Divide by

(1 + x^{2})

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

I.F. = e^{t a n^{- 1} x}

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

Put

t = e^{t a n^{- 1} x}

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

\int t d t = \frac{t ^{2}}{2}

, so

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

CBSE Class 12 Board · 2014 · Subjective