#### Provide solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 35 textbook solution.

Answer : $x y^{-1}=2 y+C, C=0$

Hint : To solve this equation we use $\frac{dy}{dx}+Py=Q$ where P,Q are constants.

Give : $\left(x+2 y^{2}\right) \frac{d y}{d x}=y \text { when } x=2, y=1$

Solution :

\begin{aligned} &\frac{d x}{d y} y=x+2 y^{2} \\ &=\frac{d x}{d y}=\frac{x+2 y^{2}}{y} \\ &=\frac{d x}{d y}=\frac{x}{y}+\frac{2 y^{2}}{y} \\ &=\frac{d x}{d y}-\frac{x}{y}=2 y \\ &=\frac{d y}{d x}+P y=Q \\ &P=-\frac{1}{y^{\prime}}, Q=2 y \end{aligned}

If  of differential equation is

\begin{aligned} &I f=e^{\int P d y} \\ &=e^{\int-\frac{1}{y} d y} \\ &=e^{-\log y} \\ &=\frac{1}{5} \end{aligned}

\begin{aligned} &x I f=\int \text { QIf } d x+C \\ &=x\left(\frac{1}{y}\right)=\int 2 y\left(\frac{1}{y}\right) d y+C \\ &=\frac{x}{y}=2 \int d y+C \\ &=\frac{x}{y}=2 y+C \end{aligned}

\begin{aligned} &=x=y(2 y+C) \\ &=x=2 y^{2}+C \\ \text { When } x &=2, y=1 \\ &=2=2+C \\ &=C=2-2=0 \end{aligned}