If y=500e^ 7x +600e^ -7x prove that d^2y dx^2 =49y.

Question

If $\text{y}=500\text{e}^{7\text{x}}+600\text{e}^{-7\text{x}}$ prove that $\frac{\text{d}^2\text{y}}{\text{dx}^2}=49\text{y}.$

✓

Answer

It is given that, $\text{y}=500\text{e}^{7\text{x}}+600\text{e}^{-7\text{x}}$
Then
$\frac{\text{dy}}{\text{dx}}=500.\frac{\text{d}}{\text{dx}}(\text{e}^{7\text{x}})+600.\frac{\text{d}}{\text{dx}}(\text{e}^{-7\text{x}})$
$=500.\text{e}^{7\text{x}}.\frac{\text{d}}{\text{dx}}(7\text{x})+600.\text{e}^{-7\text{x}}.\frac{\text{d}}{\text{dx}}(-7\text{x})$
$=3500\text{e}^{7\text{x}}-4200\text{e}^{-7\text{x}}$
$\therefore\frac{\text{d}^2\text{y}}{\text{dx}^2}=3500.\frac{\text{d}}{\text{dx}}(\text{e}^{7\text{x}})-4200.\frac{\text{d}}{\text{dx}}({-7\text{x}})$
$=3500.e^{7\text{x}}.\frac{\text{d}}{\text{dx}}(7\text{x})-4200.\text{e}^{-7\text{x}}.\frac{\text{d}}{\text{dx}}(-7\text{x})$
$=7\times3500.\text{e}^{7\text{x}}+7\times4200.\text{e}^{-7\text{x}}$
$=49\times500\text{e}^{7\text{x}}+49\times600\text{e}^{-7\text{x}}$
$=49(500\text{e}^{7\text{x}}+600\text{e}^{-7\text{x}})$
$=49\text{y}$
Hence proved

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