#### Please solve RD Sharma class 12 chapter Differential Equation exercise 21.3 question 5 maths textbook solution

$y=A \cos 2 x-B \sin 2 x$  is a solution  of differential equation.

Hint:

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

Given:

$y=A \cos 2 x-B \sin 2 x$  is a solution of differential equation.

Solution:

Differentiating on both sides with respect to $x$

$\frac{d y}{d x}=\frac{d}{d x}(A \cos 2 x-B \sin 2 x) \quad\left[\because \frac{d(u v)}{d x}=u \frac{d v}{d x}+v \frac{d u}{d x}\right]$

$\frac{d y}{d x}=\left[A \frac{d(\cos 2 x)}{d x}+\cos 2 x \frac{d A}{d x}\right]-\left[B \frac{d(\sin 2 x)}{d x}+\sin 2 x \frac{d B}{d x}\right]$

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

\begin{aligned} &\frac{d^{2} y}{d x^{2}}=-2[2 A \cos 2 x+\sin 2 x(0)]-2[-2 B \sin 2 x+\cos 2 x(0)] \\\\ &\frac{d^{2} y}{d x^{2}}=-4 A \cos 2 x+4 B \sin 2 x \end{aligned}..............(ii)

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

\begin{aligned} &\frac{d^{2} y}{d x^{2}}+4 y=0 \\\\ &L H S=\frac{d^{2} y}{d x^{2}}+4 y \end{aligned}

\begin{aligned} &=(-4 A \cos 2 x+4 B \sin 2 x)+4(A \cos 2 x-B \sin 2 x) \\\\ &=0 \\\\ &=R H S \end{aligned}

Thus, $y=A \cos 2 x-B \sin 2 x$  is a solution of differential equation.