#### Provide solution for RD Sharma class 12 chapter 21 Diffrential Equation excercise 21.4 question 3

Answer: $y=\sin x$ is the solution of given function.

Hint:

Take derivative of the function and check that the solution is satisfying or not.

Given:

$y=\sin x$ is the function.

Solution:

Differentiate with respect to x

\begin{aligned} &\Rightarrow \frac{d y}{d x}=\frac{d}{d x}(\sin x)\\ &\Rightarrow \frac{d y}{d x}=\cos x \cdots(i)\\ &\text { Differentiating eq(i) }\\ &\Rightarrow \frac{d^{2} y}{d x^{2}}=-\sin x\\ &\left.\Rightarrow \frac{d^{2} y}{d x^{2}}=-y \quad \because \because y=\sin x \text { is given }\right]\\ &\Rightarrow \frac{d^{2} y}{d x^{2}}+y=0 \end{aligned}

Thus , $y=\sin x$ satisfies the initial value problem

Now,

\begin{aligned} y(0) &=0 \\ x &=0 \\ \text { then } y &=\sin 0=0 \end{aligned}

Thus  $y(0)=0$ also satisfies initial value problem

\begin{aligned} &\text { Now, }\\ &\begin{aligned} y^{\prime}(0) &=1 \\ i e, y^{\prime} &=\cos x \\ & \therefore x=0, y^{\prime}=\cos 0=1 \end{aligned} \end{aligned}

Thus  $y^{\prime}(0)=1$ also satisfies initial value problem