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CBSE Class 12 · 2009 · Differential EquationsLong Answer

Solve the following differential equation:

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

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

Solution

Divide by

x lo g x

:

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

I.F. = e^{\int \frac{d x}{x l o g x}} = e^{l o g (l o g x)} = lo g x

y lo g x = \int \frac{2}{x ^{2}} lo g x d x

Integrate by parts (

u = lo g x

,

d v = 2 x^{- 2} d x

):

= - \frac{2 lo g x}{x} + \int \frac{2}{x ^{2}} d x = - \frac{2 lo g x}{x} - \frac{2}{x} + C

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

CBSE Class 12 Board · 2009 · Subjective

Practise more: all CBSE board PYQs on Differential Equations →

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