#### Please solve RD  sharma class 12 Chapter 21 Diffrential Equations  excercise 21.4 question 6  maths textbook solution

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

Hint:

Differentiate the function and then place value of x as per given.

Given:

$y=\sin x+\cos x$  is the function.

Solution:

Differentiating on both sides with respect to x

$\Rightarrow \frac{d y}{d x}=\cos x-\sin x \cdots(i)$

\begin{aligned} &\text { Differentiating eq(i) w.r.t } x\\ &\Rightarrow \frac{d^{2} y}{d x^{2}}=-\sin x-\cos x\\ &\Rightarrow \frac{d^{2} y}{d x^{2}}=-(\sin x+\cos x)\\ &\Rightarrow \frac{d^{2} y}{d x^{2}}=-y\\ &\Rightarrow \frac{d^{2} y}{d x^{2}}+y=0 \end{aligned}

Thus, $y=\sin x+\cos x$  satistfies the equation

\begin{aligned} &\text { When, } x=0\\ &\begin{aligned} y &=\sin 0+\cos 0 \\ &=0+1 \\ &=1 \end{aligned} \end{aligned}

$when, x=0.$

\begin{aligned} y^{\prime} &=\sin 0-\cos 0 \\ &=-1 \end{aligned}

$\therefore \text { For both } y(0)=1 \text { and } y^{\prime}(0)=-1 \text { the functions satisfies the initial value problem }$