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प्रश्न
Which of the following equations has `y = c_1e^x + c_2e^-x` as the general solution?
पर्याय
`(d^2y)/(dx^2) + y` = 0
`(d^2y)/(dx^2) - y` = 0
`(d^2y)/(dx^2) + 1` = 0
`(d^2y)/(dx^2) - 1` = 0
MCQ
उत्तर
`(d^2y)/(dx^2) - y` = 0
Explanation:
Family of curvels is `y = c_1e^x + c_2e^-x` ......(1)
Differentiating, `y^1 = c_1e^x - c_2e^-x`
`y^11 = c_1e^x + c_2e^-x = y`
∴ `y^11 - y` = 0
Solution is `(d^2y)/(dx^2) - y` = 0
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