#### Explain solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 66 Subquestion (iii) textbook solution.

Answer : $c y=\log \left|\frac{y}{x}\right|-1$

Hint : You must know the rules of solving differential equation and integration

Given : $y d x+x \log \left(\frac{y}{x}\right) d y-2 x d y=0$

Solution : $y d x+x \log \left(\frac{y}{x}\right) d y-2 x d y=0$

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

Step : 2

Putting $f(x, y)=\frac{d y}{d x} \text { and finding }(f(\lambda x, \lambda x))$

$\begin{gathered} f(x, y)=\frac{\frac{y}{x}}{2-\log \left(\frac{y}{x}\right)} \\ \qquad \begin{array}{c} f(\lambda x, \lambda y)=\frac{\frac{\lambda y}{\lambda x}}{2-\log \left(\frac{\lambda y}{\lambda x}\right)} \\ =\frac{\frac{y}{x}}{2-\log \left(\frac{y}{x}\right)} \\ f(\lambda x, \lambda y)=f(x, y) \end{array} \end{gathered}$

Step : 3

Solve by $y=vx$

Put $y=vx$

Diff w.r.t.x

\begin{aligned} &\frac{d y}{d x}=x \frac{d v}{d x}+v \frac{d x}{d x} \\ &\frac{d y}{d x}=x \frac{d v}{d x}+v \end{aligned}

Put value of $\frac{d y}{d x} \text { and } y=v x \text { in eqn(i) }$

\begin{aligned} &\frac{d y}{d x}=\frac{\frac{y}{x}}{2-\log \frac{y}{x}} \\ &x \frac{d v}{d x}=\frac{v}{2-\log v}-v \\ &x \frac{d v}{d x}=\frac{v-2 v+v \log v}{2-\log v} \end{aligned}

\begin{aligned} &x \frac{d v}{d x}=\frac{v \log v-v}{2-\log v} \\ &\frac{2-\log v}{v \log v-v} d v=\frac{d x}{x} \end{aligned}

Integrate both sides

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

Put $t=\log v-1$

$dt = \frac{1}{v}dv$

So our equation

$\int \frac{d t}{t}-\log v=\log x+\log c$

Put value of t

\begin{aligned} &\log (\log v-1)-\log v=\log x+\log c \\ &\log (\log v-1)=\log x+\log c+\log v \\ &\log (\log v-1)=\log (x c v) \\ &\text { put } v=\frac{y}{x} \end{aligned}

\begin{aligned} &\log \left(\log \frac{y}{x}-1\right)=\log x c \frac{y}{x} \\ &\log \left(\log \frac{y}{x}-1\right)=\log c y \\ &\log \frac{y}{x}-1=c y \\ &\Rightarrow c y=\log \frac{y}{x}-1 \end{aligned}