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Explain solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 66 Subquestion (v) textbook solution.

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Answer : y=\frac{x^{2}}{4}+c x^{-2}

Hint               : You must know the rules of solving differential equation and integration

Given             :     x \frac{d y}{d x}+2 y=x^{2} \quad, x \neq 0

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

Divide both sides by x

                  \begin{gathered} \frac{x}{x} \frac{d y}{d x}+\frac{2 y}{x}=\frac{x^{2}}{x} \\ \frac{d y}{d x}+\frac{2 y}{x}=x \end{gathered}

Differentiate equation in the form ,


Where, P=\frac{2}{x}, Q=x

\text { I.F }=e^{\int \frac{2}{x} d x}

\begin{array}{lc} \text { I. } F=e^{2 \log x} &\; \; \; \; \; \; \; \; \; \; \left(\log x=\log x^{n}\right) \\ \text { I. } F=e^{\log x^{2}} & \left(e^{\log x}=x\right) \end{array}

\begin{array}{lc} \text { I. } F=x^{2} \end{array}

Solution of differential equation is

\begin{gathered} y \times I . F=\int Q \times I . F d x+C \\ y x^{2}=\int x \times x^{2} d x+C \\ y x^{2}=\int x^{3} d x+C \end{gathered}

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

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