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Need solution for  RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Multiple Choice Question Question 14 textbook solution.

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Answer : (b) y^{\prime}=C_{2} e^{C_{1} x}

Hint: Variable separable method.

Given: y_{1} y_{3}=y_{2}^{2}

Explanation : Let y_{1}=y^{\prime}, y_{2}=y^{\prime \prime}, y_{3}=y^{\prime \prime \prime}

So, we rewrite (i) as

\begin{aligned} &y^{\prime} y^{\prime \prime \prime}=\left(y^{\prime \prime}\right)^{2} \\ &\frac{y^{\prime \prime \prime}}{y^{\prime \prime}}=\frac{y^{\prime \prime}}{y^{\prime}} \end{aligned}

Integrate both sides with respect to x

\begin{aligned} &\int \frac{y^{\prime \prime \prime}}{y^{\prime \prime}} d x=\int \frac{y^{\prime \prime}}{y^{\prime}} d x \\ &\log y^{\prime \prime}=\log y^{\prime}+\log C_{1} \\ &y^{\prime \prime}=C_{1} y^{\prime} \\ &\frac{y^{\prime \prime}}{y^{\prime}}=C_{1} \end{aligned}

Again integrate both sides with respect to x

\begin{aligned} &\int \frac{y^{\prime \prime}}{y^{\prime}} d x=C_{1} \int d x \\ &\log y^{\prime}=C_{1} x+\log C_{2} \\ &\log y^{\prime}-\log C_{2}=C_{1} x \end{aligned}

\begin{aligned} &\log \frac{y^{\prime}}{C_{2}}=C_{1} x \\ &y^{\prime}=C_{2} e^{c_{1} x} \end{aligned}

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