Show that y=e^ -x +ax+b is solution of the differential equation … - Vidyadip

Question

Show that $\text{y}=\text{e}^{-\text{x}}+\text{ax}+\text{b}$ is solution of the differential equation $\text{e}^\text{x}\frac{\text{d}^2\text{y}}{\text{dx}^2}=1$

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Answer

We have
$\text{y}=\text{e}^{-\text{x}}+\text{ax}+\text{b}\ ...(1)$
Differentiating both sides of (1) with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=-\text{e}^{-\text{x}}+\text{a}\ ...(2)$
Differentiating both sides of (1) with respect to x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{e}^{-\text{x}}$
$\Rightarrow\text{e}^\text{x}\frac{\text{d}^2\text{y}}{\text{dx}^2}=1$
Hence, the given function is the solution to the given differential equation.

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