Solve the following initial value problems: x dy dx -y=(x+1)e^ x …

Question

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

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Answer

We have,
$\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=(\text{x}+1)\text{e}^{\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}-\frac{1}{\text{x}}\text{y}=\Big(\frac{\text{x}+1}{\text{x}}\Big)\text{e}^{\text{x}}\ ....(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
Where $\text{P}=-\frac{1}{\text{x}}$ and $\text{Q}=\frac{\text{x}+1}{\text{x}}\text{e}^{-\text{x}}$
$\therefore\text{ I.F.}=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{-\int\frac{1}{\text{x}}\text{dx}}$
$=\text{e}^{\log\text{x}}$
$=\frac{1}{\text{x}}$
Multiplying both sides of (1) by $\text{I.F.}=\frac{1}{\text{x}},$ we get
$\frac{1}{\text{x}}\Big(\frac{\text{dy}}{\text{dx}}-\frac{1}{\text{x}}\text{y}\Big)=\Big(\frac{\text{x}+1}{\text{x}^2}\Big)\text{e}^{-\text{x}}$
Integrating both sides with respect to x, we get
$\frac{1}{\text{x}}\text{y}=\int\Big(\frac{1}{\text{x}}+\frac{1}{\text{x}^2}\Big)\text{e}^{-\text{x}}\text{dx}+\text{C}$
Putting $\frac{1}{\text{x}}\text{e}^{-\text{x}}=\text{t}$
$\Rightarrow\Big(-\frac{1}{\text{x}}\text{e}^{-\text{x}}-\frac{1}{\text{x}^2}\text{e}^{-\text{x}}\Big)\text{dx}=\text{dt}$
$\Rightarrow\Big(\frac{1}{\text{x}}+\frac{1}{\text{x}^2}\Big)\text{e}^{-\text{x}}\text{dx}=-\text{dt}$
$\therefore\ \frac{1}{\text{x}}\text{y}=\int-\text{dt}+\text{C}$
$\Rightarrow\frac{\text{y}}{\text{x}}=-\text{t}+\text{C}$
$\Rightarrow\frac{\text{y}}{\text{x}}=-\frac{\text{e}^{\text{x}}}{\text{x}}+\text{C}$
$\Rightarrow\text{y}=-\text{e}^{-\text{x}}+\text{Cx}\ ...(2)$
Now,
$\text{y}(1)=0$
$\therefore0=-\text{e}^{-1}+\text{C}$
$\Rightarrow\text{y}=\text{xe}^{-1}-\text{e}^{-\text{x}}$
Hence, $\text{y}=\text{xe}^{-1}-\text{e}^{-\text{x}}$ is the required solution.