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

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

Answers (1)

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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