Solve the following initial value problems: y'+y=e^ x , y(0)=1 2 - Vidyadip

Question

Solve the following initial value problems: $\text{y}'+\text{y}=\text{e}^{\text{x}},\text{ y}(0)=\frac{1}{2}$

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Answer

We have,
$\text{y}'+\text{y}=\text{e}^{\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}+\text{y}=\text{e}^{\text{x}}\ ...(\text{1})$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
Where $P = 1$ and $Q = e^x$
$\therefore\ \text{I.F.}=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int1\text{dx}}$
$=\text{e}^{\text{x}}$
Multiplying both sides of $(1)$ by $I.F. = e^x,$ we get
$\text{e}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}\Big)=\text{e}^{\text{x}}\text{e}^{\text{x}}$
$\Rightarrow\text{e}^{\text{x}}\frac{\text{dy}}{\text{dx}}+\text{e}^{\text{x}}\text{y}=\text{e}^{2\text{x}}$
Integrating both sides with respect to $x,$ we get
$\text{y}\text{e}^{\text{x}}=\int\text{e}^{2\text{x}}\text{dx}+\text{C}$
$\Rightarrow\text{y}\text{e}^{\text{x}}=\frac{\text{e}^{2\text{x}}}{2}+\text{C}\ ...(\text{ii})$
Now,
$\text{y}(0)=\frac{1}{2}$
$\therefore\ \frac{1}{2}\text{e}^0=\frac{\text{e}^0}{2}+\text{C}$
$\Rightarrow\text{C}=0$
Putting the value of $C$ in $(2),$ we get
$\text{y}\text{e}^{\text{x}}=\frac{\text{e}^{2\text{x}}}{2}$
$\Rightarrow\text{e}^{\text{x}}=\frac{\text{e}^{\text{x}}}{2}$
Hence, $\text{y}=\frac{\text{e}^{\text{x}}}{2}$ is the required solution.

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