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Question
Show that y = ae–3x + b, where a and b are arbitrary constants, is a solution of the differential equation `("d"^2y)/("d"x^2) + 3("d"y)/("d"x)` = 0
Solution
Given y = ae–3x + b .......(1)
Differentiating equation (1) w.r.t ‘x’, we get
`("d"y)/("d"x) = "ae"^(-33x) (- 3) + 0`
`("d"y)/("d"x) = "ae"^(- 3x) (- 3)` ........`["ae"^(- 3x) = - 1/3 ("d"y)/("d"x)]`
Again differentiating, we get
`("d"^2y)/("d"x^2) = "ae"^(-3x) (+ 9)`
`("d"^2y)/("d"x^2) = - 1/3 ("d"y)/("d"x) xx 9`
`("d"^2y)/("d"x^2) = - 3 ("d"y)/("d"x)`
`("d"^2y)/("d"x^2) + 3 ("d"y)/("d"x)` = 0
Therefore, y = ae–3x + b is a solution of the given differential equation.
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