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