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Please Solve RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 23 Maths Textbook Solution.

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Answer:  y=\left(\frac{y+1}{y}\right)+C e^{\frac{1}{y}}

Hint: To solve this equation we use    formula.

Give:  \begin{aligned} &y^{2} \frac{d x}{d y}+x-\frac{1}{y}=0 \\ & \end{aligned}

Solution:  \frac{1}{y^{2}}\left[y^{2}\left(\frac{d x}{d y}\right)+x-\frac{1}{y}\right]=0

\begin{aligned} &=\frac{d y}{d x}+\frac{x^{2}}{y^{2}}-\frac{1}{y^{3}}=0 \\ & \end{aligned}

=\frac{d y}{d x}+\frac{x^{2}}{y^{2}}=\frac{1}{y^{3}} \\

=\frac{d x}{d y}+P(x)=Q \\

P=\frac{1}{y^{\prime}} Q=\frac{1}{y^{3}} \\

I f=e^{\int P d y}

\begin{aligned} &=x I f=\int I f Q d y \\ & \end{aligned}

\frac{d x}{d y}+\frac{x}{y^{2}}=\frac{1}{y^{3}} \\

=I f=e^{\int \frac{1}{y^{2}} d y} \\

=e^{-\frac{1}{y}} \\

=x e^{-\frac{1}{y}}=e^{-\frac{1}{y}} \frac{1}{y^{3}} d y

\begin{aligned} &=\int \frac{1}{y} e^{-\frac{1}{y}} \frac{1}{y^{2}} d y \\ & \end{aligned}

=-\frac{1}{y}=t \\

=\frac{1}{y^{2}} d y=d t \\

=d y=y^{2} d t \\

-\int t e^{t} d t

\begin{aligned} &=x e^{-\frac{1}{y}}=\left[t e^{t}-e^{t}\right]+C \\ & \end{aligned}

=x e^{-\frac{1}{y}}=-t e^{t}+e^{t}+C \\

=x e^{-\frac{1}{y}}=\frac{1}{y} e^{-\frac{1}{y}}+e^{-\frac{1}{y}}+C \\

=x e^{-\frac{1}{y}}=e^{-\frac{1}{y}}\left(\frac{1}{y}+1+\frac{C}{e^{-\frac{1}{y}}}\right) \\

=x=\frac{1+y}{y}+C e^{\frac{1}{5}} 

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