Differential equation dy dx +y=0,y(0)=3 Function y=e^-x+2 - Vidyadip

Question

Differential equation $\frac{\text{dy}}{\text{dx}}+\text{y}=0,\text{y}(0)=3$

Function $\text{y}=\text{e}^\text{-x}+2$

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Answer

Here, $\text{y}=\text{e}^{\text{x}}+1 ....(1)$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}}$
$\frac{\text{dy}}{\text{dx}}=\text{y}-1 ...(2)$
Again, differentiating it with respect to x,
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=\frac{\text{dy}}{\text{dx}}$
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\frac{\text{dy}}{\text{dx}}=0$
It is given differential equation. so,
y = e^x+ 1 is a solution of the equation
put x - 0 in equation (1),
⇒ y = e⁰+ 1 = 2
y(0) = 2
put x = 0 in equation (2),
y' = e⁰ = 1
y(0) = 1

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