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Question
Show that y = e–x + mx + n is a solution of the differential equation `"e"^x(("d"^2y)/("d"x^2)) - 1` = 0
Solution
Given y = e–x + mx + n .......(1)
Differentiating equation (1) w.r.t ‘x’, we get
`("d"y)/("d"x) = "e^-x ( 1) + "m"`
`("d"y)/("d"x) = - "e"^-x + "m"`
Again differentiating, we get
`("d"^2y)/("d"x^2)= + e"^-x + 0`
`("d"^2y)/("d"x^2) = "e"-x`
Substituting the value of `("d"^2y)/("d"x^2)` in the given differential equation, we get
`"e"^x(("d"^2y)/("d"x^2)) - 1`
= ex(e-x) – 1
= en-1
= 1 – 1
= 0
Hence y = e–x + mx + n is the solution of the given differential equation `"e"^x ("d"^2y)/("d"x^2) - 1` = 0
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