If y-500 e 7 x -600 e -7 x- show that d2 y2-d x2-49 y-

Given, y=500e^7x+600e^ 7x .......(i) Differentiating twicely w.r.t. x, we get dy/dx=500e^7xd/dx(7x)+600e^ 7xd/dx( 7x)=500e^7x.7+600e^ 7x.( 7) and d^2 ...

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Standard XII

Mathematics

Chain Rule of Differentiation

If y=500 e 7 ...

Question

$I f y = 500 e^{7 x} + 600 e^{- 7 x}, s h o w t h a t \frac{d^{2} y}{d x^{2}} = 49 y .$

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Solution

Given, $y = 500 e^{7 x} + 600 e^{- 7 x}$ .......(i)

Differentiating twicely w.r.t. x, we get

$\frac{d y}{d x} = 500 e^{7 x} \frac{d}{d x} (7 x) + 600 e^{- 7 x} \frac{d}{d x} (- 7 x) = 500 e^{7 x} .7 + 600 e^{- 7 x} . (- 7) a n d \frac{d^{2} y}{d x^{2}} = (7 \times 500) e^{7 x} .7 - (7 \times 600) e^{- 7 x} (- 7) = 49 {500 e^{7 x} + 600 e^{- 7 x}} \Rightarrow \frac{d^{2} y}{d x^{2}} = 49 y . H e n c e p r o v e d .$

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If y=500e7x+600e−7x, show that d2ydx2=49 y.

Given, y=500e7x+600e−7x .......(i) Differentiating twicely w.r.t. x, we get dydx=500e7xddx(7x)+600e−7xddx(−7x)=500e7x.7+600e−7x.(−7)and d2ydx2=(7×500)e7x.7−(7×600)e−7x(−7)=49{500e7x+600e−7x}⇒ d2ydx2=49 y. Hence proved.

$I f y = 500 e^{7 x} + 600 e^{- 7 x}, s h o w t h a t \frac{d^{2} y}{d x^{2}} = 49 y .$