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

Question

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

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Answer

Given: $y = 500e^{7x} + 600e^{-7x} ...(i)$
$\therefore \frac{{dy}}{{dx}} = 500e^{7x}(7) + 600e^{-7x}(-7) = 500(7)e^{7x} - 600(7)e^{-7x}$
$\Rightarrow \frac{{{d^2}y}}{{d{x^2}}} $
$= 500(7)e^{7x}(7) - 600(7)e^{-7x}(-7) = 500(49)e^{7x} + 600(49){e^{-7x}}$
$\Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = 49[500e^{7x} + 600e^{-7x}]$
$= 49y [$From eq. $(i)]$
$\Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = 49y$ Hence proved.

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