Solve the following initial value problems: dy dx +2y=e^ -2x , y(… - Vidyadip

Question

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

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Answer

We have,
$\frac{\text{dy}}{\text{dx}}+2\text{y}=\text{e}^{-2\text{x}}\sin\text{x}\ ...(\text{i})$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
Where $\text{P}=2$ and $\text{Q}=\text{e}^{-2\text{x}}\sin\text{x}$
$\therefore\text{ I.F.}=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int2\text{dx}}$
$=\text{e}^{2\text{x}}$
Multiplying both sides of (1) by $\text{I.F.}=\text{e}^{2\text{x}},$ we get
$\text{e}^{2\text{x}}\Big(\frac{\text{dy}}{\text{dx}}+2\text{y}\Big)=\text{e}^{2\text{x}}\text{e}^-{2\text{x}}\sin\text{x}$
$\Rightarrow\text{e}^{2\text{x}}\Big(\frac{\text{dy}}{\text{dx}}+2\text{y}\Big)=\sin\text{x}$
Integrating both sides with respect to x, we get
$\text{y}\text{e}^{2\text{x}}=\int\sin\text{x dx}+\text{C}$
$\Rightarrow\text{y}\text{e}^{2\text{x}}=-\cos\text{x}+\text{C}\ ....(\text{ii})$
Now,
$\text{y}(0)=0$
$\therefore\ 0\times\text{e}^0=-\cos0+\text{C}$
$\Rightarrow\text{C}=1$
Putting the value of C in (2), we get
$\text{y}\text{e}^{2\text{x}}=-\cos\text{x}+1$
$\Rightarrow\text{y}\text{e}^{2\text{x}}=1-\cos\text{x}$
Hence, $\text{y}\text{e}^{2\text{x}}=1-\cos\text{x}$ is the required solution.

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