#### Need solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 40 textbook solution.

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

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

Hint : Using $\int \frac{1}{x} d x$

Explanation : $\left(y+3 x^{2}\right) \frac{d x}{d y}=x$

$=x \frac{d y}{d x}=y+3 x^{2}$

Divide by x

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

\begin{aligned} &=\frac{d y}{d x}-\frac{y}{x}=3 x \\ &=\frac{d y}{d x}+\left(-\frac{1}{x}\right) y=3 x \end{aligned}

This is a linear differential equation of the form

\begin{aligned} &\frac{d x}{d y}+P x=Q \\ &P=-\frac{1}{x} \text { and } Q=3 x \end{aligned}

The integrating factor $If$  of this differential equation is

\begin{aligned} &I f=e^{\int P d x} \\ &=e^{\int-\frac{1}{x} d x} \\ &=e^{-\int \frac{1}{x} d x} \\ &=e^{-\log x}\; \; \; \; \; \; \; \; \; \quad\left[\int \frac{1}{x} d x=\log x+C\right] \end{aligned}

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

Hence, the solution is

\begin{aligned} &y I f=\int Q I f d x+C \\ &=y\left(\frac{1}{x}\right)=\int 3 x\left(\frac{1}{x}\right) d x+C \\ &=\frac{y}{x}=\int 3 d x+C \end{aligned}

\begin{aligned} &=\frac{y}{x}=3 \int d x+C \\ &=\frac{y}{x}=3 x+C \; \; \; \; \; \; \; \; \quad \quad\left[\int d x=x+C\right] \end{aligned}

Multiply by x

$y=3 x^{2}+C x$

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