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

Answer : $y=x^{4}+x^{2} \log x+C x^{2}$

Hint : To solve this we convert $\text { If } \int Q I f$formula.

Give : $x d y=\left(2 y+2 x^{4}+x^{2}\right) d x$

Solution : $\frac{d y}{d x}=\frac{2 y+2 x^{4}+x^{2}}{x}$

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

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

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

\begin{aligned} &P=-\frac{2}{x^{\prime}} Q=2 x^{3}+x \\ \end{aligned}

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

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

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

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

\begin{aligned} &=\frac{1}{x^{2}} \; \; \; \; \; \; \; \; \quad\left[e^{\log x}=x\right] \\ \end{aligned}

\begin{aligned} &=y I f=\int Q I f d x+C \\ &=\frac{y}{x^{2}}=\int \frac{2 x^{3}+x}{x^{2}} d x+C \\ &=\frac{y}{x^{2}}=\int 2 x+\frac{1}{x} d x+C \\ \end{aligned}

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

\begin{aligned} &=y=x^{4}+x^{2} \log x+C x^{2} \end{aligned}