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

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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 \\


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