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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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