# If the solve the problem

Question:

If $y=A e^{-k t} \cos (p t+c)$, prove that $\frac{d^{2} y}{d t^{2}}+2 k \frac{d y}{d t}+n^{2} y=0$, where $n^{2}=p^{2}+k^{2}$

Solution:

Formula: -

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

(ii) $\frac{\mathrm{d}}{\mathrm{dx}} \mathrm{e}^{\mathrm{ax}}=\mathrm{ae}^{\mathrm{ax}}$

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

(iv) $\frac{d}{d x} \sin x=-\cos x$

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

(vi) parameteric forms $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$

Given: -

$\mathrm{y}=\mathrm{A} \mathrm{e}^{-\mathrm{kt}} \cos (\mathrm{pt}+\mathrm{c})$

Differentiating w.r.t t

$\frac{d y}{d t}=A\left(e^{-k t}(-\sin (p t+c) \cdot p)+(\cos (p t+c))\left(-r e^{-k t}\right)\right)$

$\Rightarrow \frac{d y}{d t}=-A p e^{-k t}(p t+c)-k A e^{-k t} \cos (p t+c)$

$\Rightarrow \frac{d y}{d t}=-A p e^{-k t} \sin (p t+c)-k y$

Differentiating w.r.t t

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dt}^{2}}=\mathrm{Apke}^{-\mathrm{kt}} \sin (\mathrm{pt}+\mathrm{c})-\mathrm{p}^{2} \mathrm{y}-2 \mathrm{ky}_{1}+\mathrm{ky}_{1}$

$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dt}^{2}}=\mathrm{Apke}^{-\mathrm{kt}} \sin (\mathrm{pt}+\mathrm{c})-\mathrm{p}^{2} \mathrm{y}-2 \mathrm{ky}_{1}-\mathrm{kApe}^{-\mathrm{kt}} \sin (\mathrm{pt}+\mathrm{c})-\mathrm{k}^{2} \mathrm{y}$

$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dt}^{2}}=-\left(\mathrm{p}^{2}+\mathrm{k}^{2}\right) \mathrm{y}-2 \mathrm{k} \frac{\mathrm{dy}}{\mathrm{dx}}$

$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dt}^{2}}+2 \mathrm{k} \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{n}^{2} \mathrm{y}=0$

Hence proved