Solve the following differential equation: x dy dx -y=(x-1)e^ x

Question

Solve the following differential equation:
$\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=(\text{x}-1)\text{e}^{\text{x}}$

✓

Answer

Here, $\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=(\text{x}-1)\text{e}^{\text{x}}$
$\frac{\text{dy}}{\text{dx}}-\frac{\text{y}}{\text{x}}=\Big(\frac{\text{x}-1}{\text{x}}\Big)\text{e}^{\text{x}}$
It is a linear differential equation. comparing the equation by,
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
$\text{P}=-\frac{1}{\text{x}},\text{Q}=\Big(\frac{\text{x}-1}{\text{x}}\Big)\text{e}^{\text{x}}$
I.F. $=\text{e}^{\int\text{Pdx}}=\text{e}^{-\int\frac{1}{\text{x}}\text{dx}}$
$\text{e}^{-\log|\text{x}|}=\frac{1}{\text{x}},\text{x}>0$
Solution of the equation is given by,
$\text{y}\times(\text{I.F.})=\int\text{Q}\times(\text{I.F.})\text{dx + C}$
$\text{y}\Big(\frac{1}{\text{x}}\Big)=\int\Big(\frac{\text{x}-1}{\text{x}}\Big)\text{e}^{\text{x}}\text{dx + C}$
$\frac{\text{y}}{\text{x}}=\int\Big(\frac{1}{\text{x}}-\frac{1}{\text{x}^2}\Big)\text{e}^{\text{x}}\text{dx + C}$
$\frac{\text{y}}{\text{x}}=\frac{1}{\text{x}}\text{e}^{\text{x}}+\text{C}$
Since $\int[\text{f(x)}+\text{f}'(\text{x})]\text{e}^{\text{x}}\text{dx}=\text{f(x)}\text{e}^{\text{x}}+\text{C}$
$\text{y}=\text{e}^{\text{x}}+\text{Cx},\text{x}>0$

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