Find the general solution of the following differential equation: $\left(1+y^{2}\right)+\left(x-e^{\tan^{-1}y}\right)\dfrac{dy}{dx}=0$

$x\,e^{\tan^{-1}y}=\dfrac{1}{2}e^{2\tan^{-1}y}+C$

CBSE Class 12 · 2016 · Differential EquationsLong Answer

Find the general solution of the following differential equation:

(1 + y^{2}) + (x - e^{t a n^{- 1} y}) \frac{d y}{d x} = 0

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

Rewrite as linear in

x

(with

y

independent):

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

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

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

Put

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

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

, so

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

CBSE Class 12 Board · 2016 · Subjective