Question

Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

y = e^x + 1  :  y″ – y′ = 0

Answer 1

We have y = e^x + 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 = e^x + 1 is a solution to the stated differentiating (1) equation.