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

Q15  If   y = 500 e ^{7x} + 600 e ^{- 7x } , show that \frac{d^2 y}{dx ^2} = 49 y
 

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Given function is
y = 500 e ^{7x} + 600 e ^{- 7x }
Now, differentiation w.r.t. x
\frac{dy}{dx}=\frac{d(500 e ^{7x} + 600 e ^{- 7x })}{dx}=7.500e^{7x}-7.600e^{-7x} =3500e^{7x}-4200e^{-7x}                 -(i)
Now, second order derivative is 
\frac{d^2y}{dx^2}=\frac{d^2(3500e^{7x}-4200e^{-7x})}{dx^2}
          = 7.3500e^{7x}-(-7).4200e^{-7x}= 24500e^{7x}+29400e^{-7x}                                                -(ii)
Now, we need to show
\frac{d^2 y}{dx ^2} = 49 y
Put the value of \frac{d^2y}{dx^2}  from equation  (ii)
 24500e^{7x}+29400e^{-7x}=49(500e^{7x}+600e^{-7x})
                                                 = 24500e^{7x}+29400e^{-7x}
 Hence, L.H.S. = R.H.S.
Hence proved
 

Posted by

Gautam harsolia

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