Question:
y = e^x +1 : yⁿ -y^' = 0

Answer:

y = e^x +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^' = e^x ...(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^'' = e^x

Substituting the values of y^' and y^'' in the given differential equation, we get the L.H.S. as:

y^'' - y^' = e^x - e^x = 0 = R.H.S.

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

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y = e^x +1 : yⁿ -y^' = 0

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Question:
y = e^x +1 : yⁿ -y^' = 0