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#### Explain solution for RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 24 maths textbook solution.

Answer : $x=2 y^{3}+C y^{-2}$

Hint : To solve this equation we use $xIf$ formula.

Give : $\left(2 x-10 y^{3}\right) \frac{d y}{d x}+y=0$

Solution : $\frac{d y}{d x}=-\frac{y}{2 x-10 y^{3}}$

\begin{aligned} &=\frac{d y}{d x}=\frac{-\left(2 x-10 y^{3}\right)}{y} \\ \end{aligned}

\begin{aligned} &=\frac{d y}{d x}=-\frac{2 x}{y}+\frac{10 y^{3}}{y} \\ \end{aligned}

\begin{aligned} &=\frac{d x}{d y}+\frac{2 x}{y}=10 y^{2} \\ \end{aligned}

\begin{aligned} &=\frac{d x}{d y}+R x=S \end{aligned}

\begin{aligned} &R=\frac{2 x}{y}, S=10 y^{2} \\ &I f=e^{\int R d y} \\ &=e^{2 \int \frac{1}{y} d y} \end{aligned}

\begin{aligned} &=e^{2 \log y} \\ \end{aligned}

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

\begin{aligned} &=x I f=\int S I f+C \\ &=x y^{2}=\int S I f+C \\ &=x y^{2}=\int 10 y^{2} y^{2}+C \\ &=x y^{2}=\int 10 y^{4}+C \\ \end{aligned}

\begin{aligned} &=x y^{2}=\frac{10 y^{5}}{5}+C \\ \end{aligned}

\begin{aligned} &=x y^{2}=2 y^{5}+C \\ \end{aligned}

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

\begin{aligned} &=x=2 y^{3}+C y^{-2} \end{aligned}