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The differential equation of y=k1ex+k2e-x is ______. - Mathematics and Statistics

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Question

The differential equation of `y = k_1e^x+ k_2 e^-x` is ______.

Options

  • `(d^2y)/dx^2 - y = 0`

  • `(d^2y)/dx^2 + dy/dx  = 0`

  • `(d^2y)/dx^2 + ydy/dx  = 0`

  • `(d^2y)/dx^2 + y  = 0`

MCQ
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Solution

The differential equation of `y = k_1e^x+ k_2 e^-x` is `underlinebb((d^2y)/dx^2 - y = 0)`.

Explanation:

`y = k_1e^x+ k_2 e^-x`

Differentiating w.r.t. x, we get

`dy/dx = k_1e^x -  k_2 e^-x`

Again, differentiating w.r.t. x, we get

`(d^2y)/dx^2 = k_1 e^x + k_2 e^-x`

∴ `(d^2y)/dx^2 = y`

∴ `(d^2y)/dx^2 - y = 0`

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Chapter 8: Differential Equation and Applications - Miscellaneous Exercise 8 [Page 171]

APPEARS IN

Balbharati Mathematics and Statistics 1 (Commerce) [English] Standard 12 Maharashtra State Board
Chapter 8 Differential Equation and Applications
Miscellaneous Exercise 8 | Q 1.04 | Page 171

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