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

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

Answers (1)

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 

Posted by

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