#### Explain solution for RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 23 maths textbook solution.

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

Hint : To solve this equation we use $e^{\int P d y}$ formula.

Give : $y^{2} \frac{d x}{d y}+x-\frac{1}{y}=0$

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}

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

\begin{aligned} &=\frac{d x}{d y}+P(x)=Q \\ \end{aligned}

\begin{aligned} &P=\frac{1}{y^{\prime}} Q=\frac{1}{y^{3}} \\ \end{aligned}

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

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

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

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

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

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

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

\begin{aligned} &=-\frac{1}{y}=t \\ \end{aligned}

\begin{aligned} &=\frac{1}{y^{2}} d y=d t \\ \end{aligned}

\begin{aligned} &=d y=y^{2} d t \\ \end{aligned}

\begin{aligned} &=\int t e^{t} d t \\ \end{aligned}

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

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

\begin{aligned} &=x e^{-\frac{1}{y}}=\frac{1}{y} e^{-\frac{1}{y}}+e^{-\frac{1}{y}}+C \\ \end{aligned}

\begin{aligned} &=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}} \end{aligned}