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

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

Answers (1)

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

Posted by

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