Question

Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation: y=ex+1 : y''-y'=0

Answer 1

y=ex+1
Differentiating both sides of this equation with respect to x, we get:

$\frac{dy}{dx}=\frac{d}{dx}(e^x+1)$

$\Rightarrow y'=e^x$...(1)
Now, differentiating equation (1) with respect to x we get:

$\frac{d}{dx}(y')=\frac{d}{dx}(e^x)$

$\Rightarrow y''=e^x$
Substituting the values of y' and y'' in the given differential equation, we get the L.H.S. as:
y''-y'=ex-ex=0 = R.H.S.

Thus, the given function is the solution of the corresponding differential equation.