Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation: `y=e^x+1"""""""":""yprimeprime-yprime=0`

Explanation

`y=e^{x}+1`


Differentiating equation with respect to x, we find


`frac{d y}{d x}=frac{d}{d x}(e^{x}+1)`


`Rightarrow y^{prime}=e^{x}1 `


Again differentiating equation with respect to x, we find


`frac{d}{d x}(y^{prime})=frac{d}{d x}(e^{x}) `


`Rightarrow y^{prime prime}=e^{x}`


Substituting the values of `y^{prime} and y^{prime prime}` in differential equation,


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