#### Need solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.7 question 59

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

Hint: Separate the terms of x and y and then integrate them.

Given: $\left(1-y^{2}\right)(1+\log x) d x+2 x y d y=0 \text { given that } y=0 \text { when } x=1$

Solution:

\begin{aligned} &\left(1-y^{2}\right)(1+\log x) d x=-2 x y d y \\\\ &\Rightarrow \frac{1+\log x}{x} d x=\frac{-2 y}{1-y^{2}} d y \end{aligned}

Integrating both sides

\begin{aligned} &\int \frac{1+\log x}{x} d x=\int \frac{-2 y}{1-y^{2}} d y \ldots \ldots \ldots \ldots . .(*) \\ &\Rightarrow I_{1}=I_{2} \text { where } I_{1}=\int \frac{1}{x}(1+\log x) d x \text { and } \\\\ &I_{2}=\int \frac{-2 y}{1-y^{2}} d y \\\\ &\therefore I_{1}=\int \frac{1}{x}(1+\log x) d x \end{aligned}

Put

\begin{aligned} &1+\log x=t \\\\ &\frac{1}{x} d x=d t \\\\ &=\int t d t \\\\ &=\frac{t^{2}}{2}+c=\frac{(1+\log x)^{2}}{2}+c \end{aligned}

Now,   $I_{2}=\int \frac{-2 y}{1-y^{2}} d y$

Put

\begin{aligned} &1-y^{2}=t \\\\ &-2 y d y=d t \\\\ &=\int \frac{d t}{t} \\\\ &=\log |t|+c \\\\ &=\log \left|1-y^{2}\right|+c \end{aligned}

Put the values in (*)  we get

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

Now according to given $y = 0$ when $x = 1$

\begin{aligned} &\therefore \frac{(1+\log 1)^{2}}{2}=\log |1-0|+c \\\\ &\Rightarrow \frac{(1+0)^{2}}{2}=\log (1)+c \\\\ &{[\log 1=0]} \end{aligned}

\begin{aligned} &\Rightarrow 1=2[(0)+c] \\\\ &\Rightarrow 1=2 c \Rightarrow c=\frac{1}{2} \end{aligned}

Put in (1) we get

\begin{aligned} &\Rightarrow \frac{(1+\log x)^{2}}{2}=\log \left|1-y^{2}\right|+\frac{1}{2} \\\\ &\Rightarrow(1+\log x)^{2}=2 \log \left(1-y^{2}\right)+1 \\\\ &\Rightarrow(1+\log x)^{2}=\log \left(1-y^{2}\right)^{2}+1 \end{aligned}