If the solve the problem

Question:

If $y=500 e^{7 x}+600 e^{-7 x}$, show that $\frac{d^{2} y}{d x^{2}}=49 y$

Solution:

Formula: -

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

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

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

(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: -

$y=500 e^{7 x}+600 e^{-7 x}$

$\frac{d y}{d x}=500 \cdot \frac{d\left(e^{7 x}\right)}{d x}+600 \cdot \frac{d\left(e^{-7 x}\right)}{d x}$

$\Rightarrow \frac{d y}{d x}=500 e^{7 x} \cdot \frac{d(7 x)}{d x}+600 \cdot e^{7 x} \cdot \frac{d(-7 x)}{d x}$

$\Rightarrow \frac{d y}{d x}=3500 e^{7 x}-4200 e^{-7 x}$

$\Rightarrow \frac{d y}{d x}=49\left(500 e^{7 x}+600 e^{-7 x}\right)$

$\Rightarrow \frac{d y}{d x}=49 y$

Hence proved.

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