#### Please Solve RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 35 Maths Textbook Solution.

Answer:  $x y^{-1}=2 y+C, C=0$

Hint: To solve this equation we use $\frac{d y}{d x}+P y=Q$  where $P,Q$  are constants.

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

Solution:  $\frac{d x}{d y} y=x+2 y^{2}$

\begin{aligned} &=\frac{d x}{d y}=\frac{x+2 y^{2}}{y} \\ & \end{aligned}

$=\frac{d x}{d y}=\frac{x}{y}+\frac{2 y^{2}}{y} \\$

$=\frac{d x}{d y}-\frac{x}{y}=2 y \\$

$=\frac{d y}{d x}+P y=Q \\$

$P=-\frac{1}{y^{\prime}} Q=2 y$

$If$  of differential equation is

\begin{aligned} &I f=e^{\int P d y} \\ & \end{aligned}

$=e^{\int-\frac{1}{y} d y} \\$

$=e^{-\log y} \\$

$=\frac{1}{5} \\$

$x I f=\int \text { QIf } d x+C \\$

$=x\left(\frac{1}{y}\right)=\int 2 y\left(\frac{1}{y}\right) d y+C$

\begin{aligned} &=\frac{x}{y}=2 \int d y+C \\ & \end{aligned}

$=\frac{x}{y}=2 y+C \\$

$=x=y(2 y+C) \\$

$=x=2 y^{2}+C \\$

$\text { When } x =2, y=1 \\$

$=2=2+C \\$

$=C=2-2=0$