Question

Solve the given initial value problem. Give the largest interval lover which the general solution is defined.

${\mathbf{xy}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{y}\mathbf{=}{\mathbf{e}}^{\mathbf{x}}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{2}$

Short answer

The solution for the given initial value is $\mathrm{y}={\mathrm{x}}^{-1}{\mathrm{e}}^{\mathrm{x}}+(2-\mathrm{e}){\mathrm{x}}^{-1}$.

Step-by-step solution

The given equation in the standard form and determine the integrated factor

Consider the initial value problem,

$x{y}^{\text{'}}+y=e^x,y\left(1\right)=2$

The objective is to solve the initial value problem and also give the largest interval over which the solution is defined.

Divide the differential equation both sides with x.

$y^{\text{'}}+\frac{1}{x}y=\frac{e^x}{x}...........\left(1\right)$

Compare the equation with the standard form $y^{\text{'}}+P\left(x\right)y=f\left(x\right)$to get,

$P\left(x\right)=\frac{1}{x}\text{and}f\left(x\right)=\frac{e^x}{x}$

$\text{Observe that,}P\left(x\right)=\frac{1}{x}\text{and}f\left(x\right)=\frac{e^x}{x}$are continuous on interval $\left(0,\infty \right)$.

$$\begin{gathered} \text{Integrating factor}=e^{\int P\left(x\right)dx} \\ =e^{\int \frac{1}{x}dx} \\ =e^{\ln x} \\ =x \end{gathered}$$

Determine the general solution for the given differential equation

Multiply the differential equation (1) with integrating factor $x\left(y^{\text{'}}+\frac{1}{x}y\right)=x\frac{e^x}{x}$

$$\begin{gathered} x{y}^{\text{'}}+y=e^x \\ \frac{d}{dx}\left(xy\right)=e^x \\ xy=\int e^{x}dx \\ xy=e^x+c \\ y=\frac{e^x}{x}+\frac{c}{x}\text{for}0<x<\infty \end{gathered}$$

To find the value of the integration constant in the general solution use the initial condition $y\left(1\right)=2$.

$y\left(x\right)=\frac{e^x}{x}+\frac{c}{x}$General solution

Substitute the value of in the general solution.

$$\begin{gathered} y\left(1\right)=\frac{e^1}{1}+\frac{c}{1} \\ 2=e+c \\ c=2-e \end{gathered}$$

$$\begin{gathered} y=\frac{e^x}{x}+\frac{2-e}{x}\text{for}0<x<\infty \\ y=x^{-1}{e}^x+(2-e)x^{-1}x\in \left(0\right) \end{gathered}$$

Therefore, solution of the given initial value problem is,

$y=x^{-1}{e}^x+(2-e)x^{-1},x\in (0,\infty )$