If y = 500e7x + 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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