Find the particular solution of the differential equation $\left(1-y^{2}\right)(1+\log x)\,dx+2xy\,dy=0$, given that $y=0$ when $x=1$.

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

CBSE Class 12 · 2016 · Differential EquationsLong Answer

Find the particular solution of the differential equation

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

, given that

y = 0

when

x = 1

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

Separate variables:

\frac{1 + lo g x}{x} d x + \frac{2 y}{1 - y ^{2}} d y = 0

With

t = 1 + lo g x

in the first integral and

p = 1 - y^{2}

in the second:

\frac{( 1 + lo g x ) ^{2}}{2} - lo g ∣1 - y^{2} ∣ + C = 0

Apply

y = 0, x = 1

\frac{1}{2} - 0 + C = 0 \Rightarrow C = - \frac{1}{2}

. Therefore

(1 + lo g x)^{2} = 2 lo g ∣1 - y^{2} ∣ + 1

CBSE Class 12 Board · 2016 · Subjective