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Explain solution RD Sharma class 12 chapter Differential Equation exercise 21.3 question 4 maths

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y=A \cos x+B \sin x  is a solution of differential equation.


Differentiate the given solution of differential equation on both sides with respect to x


 y=A \cos x+B \sin x  is a solution.


Differentiating on both sides with respect to x

\frac{d y}{d x}=\frac{d}{d x}(A \cos x+B \sin x) \quad\left[\because \frac{d(u v)}{d x}=u \frac{d v}{d x}+v \frac{d u}{d x}\right]

\frac{d y}{d x}=\left[A \frac{d(\cos x)}{d x}+\cos x \frac{d A}{d x}\right]+\left[B \frac{d(\sin x)}{d x}+\sin x \frac{d B}{d x}\right]

\begin{aligned} &\frac{d y}{d x}=[-A \sin x+0]+[B \cos x+0] \\\\ &\frac{d y}{d x}=-A \sin x+B \cos x \end{aligned}        ...........(i)

Now to obtain the second order derivative, differentiate equation (i) with respect to x

\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}(-A \sin x+B \cos x) \\\\ &\frac{d^{2} y}{d x^{2}}=\left[-A \frac{d(\sin x)}{d x}+\sin x \frac{d(-A)}{d x}\right]+\left[B \frac{d(\cos x)}{d x}+\cos x \frac{d B}{d x}\right] \end{aligned}

\begin{aligned} &\frac{d^{2} y}{d x^{2}}=[-A \cos x+0]+[-B \sin x] \\\\ &\frac{d^{2} y}{d x^{2}}=-A \cos x-B \sin x \end{aligned}.................(ii)

Now put equation (ii) in given differential equation as follows

\begin{aligned} &\frac{d^{2} y}{d x^{2}}+y=0 \\\\ &L H S=\frac{d^{2} y}{d x^{2}}+y \end{aligned}

            \begin{aligned} &=(-A \cos x-B \sin x)+(A \cos x+B \sin x) \\\\ &=0 \\\\ &=R H S \end{aligned}


Thus, y=A \cos x+B \sin x is a solution of differential equation.

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