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Need Solution for R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise Revision Exercise Question 12 Maths Textbook Solution.

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Answer:  Hence verified

Hint: Find first and second derivative of given equation and put in differential equation to be verified.

Given: y=cx+2c^{2}\; \; \; \; ,2\left ( \frac{dy}{dx} \right )^{2}+x\frac{dy}{dx}-y=0

Solution:  y=C\: x+2\: c^{2}

\frac{dy}{dx}=c…differentiating w.r.t to x

\frac{d^{2}y}{dx^{2}}=0..differentiating agin w.r.t to x

Now, differential equation is

\begin{aligned} &2\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}+x \frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}=0 \\ &\mathrm{~L} . \mathrm{H} \cdot \mathrm{S}=2(\mathrm{c})^{2}+\mathrm{x}(\mathrm{c})-\mathrm{c} x-2 \mathrm{c}^{2} \\ &=2 \mathrm{c}^{2}+\mathrm{c} \mathrm{x}-\mathrm{c} \mathrm{x}-2 \mathrm{c}^{2}=0=\mathrm{R} . \mathrm{H} . \mathrm{S} \end{aligned}

Hence verified

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