Solve the following differential equation: $\left(1+x^{2}\right)\dfrac{dy}{dx}-2xy=\left(x^{2}+2\right)\left(x^{2}+1\right)$

$y=\left(x+\tan^{-1}x+C\right)\left(1+x^{2}\right)$

CBSE Class 12 · 2005 · Differential EquationsLong Answer

Solve the following differential equation:

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

Answer: $y = (x + tan^{- 1} x + C) (1 + x^{2})$

Divide by

(1 + x^{2})

to get linear form:

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

I.F. = e^{\int - \frac{2 x}{1 + x ^{2}} d x} = e^{- l o g (1 + x^{2})} = \frac{1}{1 + x ^{2}}

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

Integrate:

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

y = (x + tan^{- 1} x + C) (1 + x^{2})

CBSE Class 12 Board · 2005 · Subjective