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Question:

If $y=e^{x}(\sin x+\cos x)$ Prove that $\frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+2 y=0$

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

(iii) $\frac{\mathrm{d}}{\mathrm{dx}} \mathrm{x}^{\mathrm{n}}=\mathrm{n} \mathrm{x}^{\mathrm{n}-1}$

(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: –

$y=e^{x}(\sin x+\cos x)$

differentiating w.r.t $x$

$\frac{d y}{d x}=e^{x}(\cos x-\sin x)+(\sin x+\cos x) e^{x}$

$\Rightarrow \frac{d y}{d x}=y+e^{x}(\cos x-\sin x)$

Differentiating w.r.t $x$

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{e}^{\mathrm{x}}(-\sin \mathrm{x}-\cos \mathrm{x})+(\cos \mathrm{x}-\sin \mathrm{x}) \mathrm{e}^{\mathrm{x}}$

$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}+(\cos \mathrm{x}-\sin \mathrm{x}) \mathrm{e}^{\mathrm{x}}$

Adding and subtracting y on RHS

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}+(\cos \mathrm{x}-\sin \mathrm{x}) \mathrm{e}^{\mathrm{x}}+\mathrm{y}-\mathrm{y}$

$\Rightarrow \frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+2 y=0$

Hence proved