If the solve the problem

Question:

If $y=e^{x}(\sin x+\cos x)$ prove that $\frac{d^{2} y}{d x^{2}}-1 \frac{d y}{d x}+2 y=0$

Solution:

Formula:

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

(ii) $\frac{d\left(e^{a x}\right)}{d x}=a e^{a x}$

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

(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: -

$y=a e^{2 x}+b e^{-2 x}$

Differentiating w.r.t $x$

$\frac{d y}{d x}=2 a e^{2 x}+b e^{(-x)}(-1)$

$\Rightarrow \frac{d y}{d x}=2 a e^{2 x}-b e^{-x}$

Differentiating w.r.t x

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=2 \mathrm{ae}^{2 \mathrm{x}}(2)-\mathrm{be}^{-\mathrm{x}}(-1)$

$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=4 \mathrm{ae}^{2 \mathrm{x}}+\mathrm{be}^{-\mathrm{x}}$

Adding and subtracting be $^{-x}$ on RHS

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=4 \mathrm{ae}^{2 \mathrm{x}}+2 \mathrm{be}^{-\mathrm{x}}$

$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=2 \mathrm{y}+\frac{\mathrm{dy}}{\mathrm{dx}}$

$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}-\frac{\mathrm{dy}}{\mathrm{dx}}-2 \mathrm{y}=0$

$\Rightarrow \frac{d^{2} y}{d x^{2}}=2\left(a e^{2 x}+b e^{-x}\right)+2 a e^{2 x}-b e^{-x}$

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