#### Provide solution for RD Sharma maths class 12 chapter Higher Order Derivatives exercise 11.1 question 50

$A=\frac{2}{3}\: and\: B=\frac{-1}{3}$

Hint:

You must know the derivative of second order

Given:

$\text { Find } A \& B \text { so that } y=A \sin 3 x+B \cos 3 x \text { satisfies the equation } \frac{d^{2} y}{d x^{2}}+4 \frac{d y}{d x}+3 y=10 \cos 3 x$

Solution:

\begin{aligned} &\text { Let } y=A \sin 3 x+B \cos 3 x \\ &\frac{d y}{d x}=\frac{d(A \sin 3 x+B \cos 3 x)}{d x} \\ &=A \cos 3 x-3+b(-\sin 3 x .3) \quad\left[\begin{array}{l} \frac{d \cos 3 x}{d x}=-3 \sin 3 x \\ \frac{d \sin 3 x}{d x}=3 \cos 3 x \end{array}\right] \end{aligned}

\begin{aligned} &\frac{d y}{d x}=3 A \cos 3 x-3 B \sin 3 x \\ &\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}(3 A \cos 3 x-3 B \sin 3 x) \\ &=3 A(-\sin 3 x \cdot 3)-3 B(\cos 3 x .3) \\ &=-9 A \sin 3 x-9 B \cos 3 x \\ &=-9(A \sin 3 x+3 \cos 3 x) \\ &=-9 y \end{aligned}

\begin{aligned} &\frac{d^{2} y}{d x^{2}}+4 \frac{d y}{d x}+3 y=10 \cos 3 x\\ &-9 y+4(3 A \cos 3 x-3 B \sin 3 x)+3 y=10 \cos 3 x\\ &-6 y+12 A \cos 3 x-12 B \sin 3 x=10 \cos 3 x\\ &-6(A \sin 3 x+B \cos 3 x)+12 A \cos 3 x-12 B \sin 3 x=10 \cos 3 x\\ &\sin 3 x(-6 A-12 B)+\cos 3 x(-6 B+12 A)=10 \cos 3 x\\ &-6 A-12 B=0 \quad \ldots \ldots . .(1)\\ &-6 B+12 A=0 \quad \ldots \ldots . .(2)\\ &6 A=-12 B\\ &A=-2 B \quad \ldots \ldots(3) \end{aligned}

\begin{aligned} &\text { Put (3) in (2) }\\ &-6 B+(-2 B) 12=10\\ &-6 B-24 B=10\\ &-30 B=10\\ &B=\frac{-1}{3}\\ &A=-2\left(\frac{-1}{3}\right)=\frac{2}{3}\\ &A=\frac{2}{3}\; \&\; B=\frac{-1}{3} \end{aligned}