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Provide Solution for RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 40

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

Give:  \begin{aligned} & &\left(y+3 x^{2}\right) \frac{d x}{d y}=x \\ \end{aligned}

Hint: Using   \begin{aligned} & & \int \frac{1}{x} d x \end{aligned}

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

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

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} \end{aligned}

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

\frac{d x}{d y}+P x=Q

P=-\frac{1}{x} \text { and } Q=3 x


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

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


Multiply by x

=y=3 x^{2}+C x 

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