#### Please solve RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 66 Subquestion (xiv) textbook solution.

Answer : $x=\frac{y^{3}}{3}+\frac{c}{y}$

Hint : integrate by applying integration of $x^{n}$

Given : $y \; d x+\left(x-y^{2}\right) d y=0$

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

$\frac{d y}{d x}=\frac{-y}{x-y^{2}}$

This is not in the form of $\frac{d y}{d x}+Py=Q$

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

\begin{aligned} &\frac{d x}{d y}=y-\frac{x}{y} \\ \end{aligned}

\begin{aligned} &\frac{d x}{d y}+\frac{x}{y}=y \end{aligned}                                               .....(i)

Find P and Q Where $P=\frac{1}{y} \quad, Q=y$

Find I.F

\begin{aligned} \text { I.F } &=e^{\int P d y} \\ &=e^{\int \frac{1}{y} d y}=e^{\log y}=y \end{aligned}

Solution will be

\begin{aligned} &x(I . F)=\int(Q \times I . F) d y+c \\ &x y=\int y \times y d y+c \\ &x y=\int y^{2} d y+c \end{aligned}

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