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Question
Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:
y = ex + 1 : y″ – y′ = 0
Solution
We have y = ex + 1 ...(1)
Differentiating (1) w.r.t.x, we get
`y' = d/dx (e^x + 1) = e^x`
and `y” = d/dx (e^x) = e^x`
⇒ y” - y’ = 0
Thus, y = ex + 1 is a solution to the stated differentiating (1) equation.
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