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

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


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


y=\sin x+\cos x  is the function.


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 }

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