Question 1

# y = ex +1 : yn -y' = 0

y = ex +1

Differentiating both sides of this equation with respect to x, we get:

\begin{align}\frac{dy}{dx}=\frac{d}{dx}(e^x + 1)\end{align}

=> y' = ex                          ...(1)

Now, differentiating equation (1) with respect to x, we get:

\begin{align}\frac{d}{dx}(y^{'})=\frac{d}{dx}(e^x)\end{align}

=> y'' = ex

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.

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