# find the problem

Question:

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

Solution:

Formula: -

(i) $\frac{\mathrm{dy}}{\mathrm{dx}}=y_{1}$ and $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=y_{2}$

(ii) $\frac{\mathrm{d}}{\mathrm{dx}} \cos \mathrm{x}=\sin \mathrm{x}$

(iii) $\frac{\mathrm{d}}{\mathrm{dx}} \sin \mathrm{x}=-\cos \mathrm{x}$

(iv) chain rule $\frac{\mathrm{df}}{\mathrm{dx}}=\frac{\mathrm{d} \text { (wou) }}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}}=\frac{\mathrm{dw}}{\mathrm{ds}} \cdot \frac{\mathrm{ds}}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}}$

Given: -

$y=A \sin 3 x+B \cos 3 x$

differentiating w.r.t $x$

$\frac{d y}{d x}=3 \operatorname{Axos} 3 x+3 B(-\sin 3 x)$

Again differentiating w.r.t $x$

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=3 \mathrm{~A}(-\sin 3 \mathrm{x}) \cdot 3-3 \mathrm{~B}(\cos 3 \mathrm{x}) \cdot 3$

$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-9(A \sin 3 \mathrm{x}+B \cos 3 \mathrm{x})=-9 \mathrm{y}$

$\frac{d^{2} y}{d x^{2}}+\frac{4 d y}{d x}+3 y$

$\Rightarrow \frac{d^{2} y}{d x^{2}}+\frac{4 d y}{d x}+3 y=-9 y+4(3 A \cos 3 x-3 B \sin 3 x)+3 y$

$\Rightarrow \frac{d^{2} y}{d x^{2}}+\frac{4 d y}{d x}+3 y=12(A \cos 3 x-b \sin 3 x)-6(A \sin 3 x+B \cos 3 x)$

$\Rightarrow \frac{d^{2} y}{d x^{2}}+\frac{4 d y}{d x}+3 y=(12 A-6 B) \cos 3 x-(12 B+6 A) \sin 3 x$

But given,

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}+\frac{4 \mathrm{dy}}{\mathrm{dx}}+3 \mathrm{y}=10 \cos 3 \mathrm{x}$

$\Rightarrow 12 A-6 B=10$

$\Rightarrow-(12 B+6 A)=0$

$\Rightarrow 6 A=-12 B$

$\Rightarrow A=-2 B$

Puttuing A

$\Rightarrow 12(-2 B)-63=10$

$\Rightarrow-24 B-6 B=10$

$\Rightarrow \mathrm{B}=-\frac{1}{3}$

$A=-2 \times-\frac{1}{3}=\frac{2}{3}$

And $A=\frac{2}{3}, B=-\frac{1}{3}$