Question

If y=500e^7x+600e^-7x, show that d^2y/dx^2=49y

Answer 1

Given, $\bf{y=500e^{7x}+600e^{-7x}}$
we have to proof :$\bf{\frac{d^2y}{dx^2}=49y}$

y = 500e^{7x} + 600e^{-7x}-------(1)

differentiate y with respect to x,

dy/dx = 500. d(e^{7x})/dx + 600. d(e^{-7x})/dx

= 500 × 7.e^{7x} + 600 × (-7).e^{-7x}

= 7[500e^{7x} - 600e^{-7x}]

again differentiate dy/dx with respect to x,

d²y/dx² = 7{500. d(e^{7x})/dx - 600. d(e^{-7x})/dx}

= 7[500.7.e^{7x} - 600 × (-7) .e^{-7x}]

= 49[500.e^{7x} + 600.e^{-7x}]

d²y/dx² = 49y [ from equation (1),

d²y/dx² = 49y $\textbf{\underline{hence proved}}$

Answer 2

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