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CBSE Class 12 · 2011 · Differential EquationsLong Answer

Solve the following differential equation:

(1 + y^{2}) (1 + lo g x) d x + x d y = 0

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

Solution

Separate variables:

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

On the right put

z = 1 + lo g x

,

d z = \frac{d x}{x}

:

tan^{- 1} y = - \int z d z = - \frac{z ^{2}}{2} + C

tan^{- 1} y = - \frac{1}{2} (1 + lo g x)^{2} + C

CBSE Class 12 Board · 2011 · Subjective

Practise more: all CBSE board PYQs on Differential Equations →

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