Question

Question: In Problems $\mathbf{17}\mathbf{-}\mathbf{22}$, solve the initial value problem.

$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{-}\frac{\mathbf{y}}{\mathbf{x}}\mathbf{=}{\mathbf{xe}}^{\mathbf{x}}\mathbf{,}\mathbf{}\mathbf{}\mathbf{y}\left(1\right)\mathbf{=}\mathbf{e}\mathbf{-}\mathbf{1}$

Short answer

The initial value of the given equation is $\mathrm{y}={\mathrm{xe}}^{\mathrm{x}}-\mathrm{x}$.

Step-by-step solution

Method for solving linear equations

• Write the equation in the standard form $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathbf{P}\left(x\right)\mathbf{y}\mathbf{=}\mathbf{Q}\left(x\right)$.
• Calculate the integrating factor $\mathit{\mu }\left(x\right)$by the formula $\mathit{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathit{e}\mathit{x}\mathit{p}\left[{\int }_{}^{}P\left(x\right)dx\right]$.
• Multiply the equation in standard form by $\mathit{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$and, recalling that the left-hand side is just $\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\left[\mu \left(x\right)y\right]$, obtain

$$\begin{gathered} \mathit{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{+}\mathit{P}\mathit{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathit{y}\mathbf{=}\mathit{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathit{Q}\left(x\right) \\ \frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\left[\mu \left(x\right)y\right]\mathbf{=}\mathit{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathit{Q}\mathbf{\left(}\mathbf{x}\mathbf{\right)} \end{gathered}$$

Integrate the last equation and solve for y by dividing by $\mathit{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$to obtain $\mathbf{y}\left(x\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{\pi }\left(x\right)}\left[{\int }_{}^{}\pi \left(x\right)Q\left(x\right)+c\right]$. Here C is an arbitrary constant

 Step 2: Solve the given equation

Given that,

$\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{\mathrm{y}}{\mathrm{x}}={\mathrm{xe}}^{\mathrm{x}},\mathrm{y}\left(1\right)=\mathrm{e}-1......\left(1\right)$

Write the equation (1) into standard form of linear equation.

$$\begin{gathered} \frac{\mathrm{dy}}{\mathrm{dx}}-\frac{\mathrm{y}}{\mathrm{x}}={\mathrm{xe}}^{\mathrm{x}} \\ \frac{\mathrm{dy}}{\mathrm{dx}}-\frac{1}{\mathrm{x}}y={\mathrm{xe}}^{\mathrm{x}}........\left(2\right) \end{gathered}$$

Calculate the integrating factor of $\mu \left(x\right)$.

Where $\text{P}\left(\mathrm{x}\right)=-\frac{1}{\mathrm{x}}$.

Then,

role="math" localid="1664121521057" $\begin{array}{rcl}\mathrm{\mu }\left(\mathrm{x}\right)& =& {\mathrm{e}}^{{\int }_{}^{}\mathrm{P}\left(\mathrm{x}\right)\mathrm{dx}}\\ & =& {\mathrm{e}}^{{\int }_{}^{}-\frac{1}{\mathrm{x}}\mathrm{dx}}\\ & =& {\mathrm{e}}^{\left(-\ln \left|\mathrm{x}\right|\right)}\\ & =& \frac{1}{x}\end{array}$

Simplification method

Multiply $\mu \left(x\right)$in equation (2)

$$\begin{gathered} \frac{1}{\mathrm{x}}\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{1}{\mathrm{x}}\frac{1}{\mathrm{x}}\mathrm{y}=\frac{1}{\mathrm{x}}{\mathrm{xe}}^{\mathrm{x}} \\ \frac{1}{\mathrm{x}}\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{1}{{\mathrm{x}}^2}\mathrm{y}={\mathrm{e}}^{\mathrm{x}} \\ \frac{\mathrm{d}}{\mathrm{dx}}\left[\frac{1}{\mathrm{x}}\mathrm{y}\right]={\mathrm{e}}^{\mathrm{x}} \end{gathered}$$

Integrating both sides.

$$\begin{gathered} {\int }_{}^{}\frac{\mathrm{d}}{\mathrm{dx}}\left[\frac{1}{\mathrm{x}}\mathrm{y}\right]\mathrm{dx}={\int }_{}^{}{\mathrm{e}}^{\mathrm{x}}\mathrm{dx} \\ \frac{1}{\mathrm{x}}\mathrm{y}={\mathrm{e}}^{\mathrm{x}}+\mathrm{C} \\ \mathrm{y}={\mathrm{xe}}^{\mathrm{x}}+\mathrm{Cx}.........\left(3\right) \end{gathered}$$

Find the initial value

Given that, $y\left(1\right)=e-1$.

Then, $x=1$and $y=e-1$.

Substitute the value in equation (3) to get the value of C.

$\begin{array}{rcl}\mathrm{y}& =& {\mathrm{xe}}^{\mathrm{x}}+\mathrm{Cx}\\ \mathrm{e}-1& =& \mathrm{e}+\mathrm{C}\\ -1& =& \mathrm{C}\end{array}$

Substitute the value of C in equation (2).

$\mathrm{y}={\mathrm{xe}}^{\mathrm{x}}-\mathrm{x}$

So, the solution is$\mathbf{y}\mathbf{=}\mathit{x}{\mathit{e}}^{\mathbf{x}}\mathbf{-}\mathbf{x}$