#### Provide solution for  RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 58 textbook solution.

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

Hint                : you must know the rules of solving differential equation and integration

Given             $y^{2}+\left(x+\frac{1}{y}\right) \frac{d y}{d x}=0$

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

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

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

Where  , $P=\frac{1}{y^{2}}, Q=\frac{-1}{y^{3}}$

Now,

$\text { I. } F=e^{\int \frac{1}{y^{2}} d y}=e^{\frac{-1}{y}}$

therefore the solution is

\begin{aligned} x \times I . F &=\int \text { I.F } \times Q d y+c \\ x e^{\frac{-1}{y}} &=\int-e^{\frac{-1}{y}} \frac{1}{y^{3}} d y+c \\ x e^{\frac{-1}{y}} &=I+c \end{aligned}

Putting $t=\frac{1}{y},$ and differentiating both sides

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

Applying integration both side,

\begin{aligned} I &=\int\left(t \times e^{-t}\right) d t \\ I &=t \times \int e^{-t} d t-\int\left(\frac{d t}{d t} \times \int e^{-t} d t\right) d t \\ &=-t e^{-t}+\int e^{-t} d t \\ &=-t e^{-t}-e^{-t} \end{aligned}

Therefore,

\begin{aligned} &I=\frac{-1}{y} e^{\frac{-1}{y}}-e^{\frac{-1}{y}} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[t=\frac{1}{y}\right] \\ &I=-e^{\frac{-1}{y}}\left(1+\frac{1}{y}\right) \end{aligned}

Hence, required solution is ,

\begin{aligned} &x e^{\frac{-1}{y}}=-e^{\frac{-1}{y}}\left(1+\frac{1}{y}\right)+c \\ &x=-\left(1+\frac{1}{y}\right)+c e^{\frac{1}{y}} \end{aligned}