Question

Find the general solution of the given differential equation. Give the largest interval /over which the general solution is defined. Determine whether there are any transient terms in the general solution.

${\mathbf{xy}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{x}\mathbf{)}\mathbf{y}\mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{x}}\mathbf{sin}\mathbf{2}\mathbf{x}$

Short answer

The general solution for differential equation is $\mathrm{y}=-\frac{1}{2\mathrm{x}}{\mathrm{e}}^{-\mathrm{x}}\cos 2\mathrm{x}+\frac{{\mathrm{Ce}}^{-\mathrm{x}}}{\mathrm{x}}$and

there is transient term in the general solution.

Step-by-step solution

Determine the linear differential equation in the variable  y

Consider the differential equation,

$x{y}^{\text{'}}+(1+x)y=e^{-x}\sin 2x$

The objective is to find the general solution of above differential equation and find the largest interval l over which the general solution is defined.

Divide both sides by x to get standard form of linear differential equation.

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

Thus, the linear differential equation in the variable y is,

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

Determine the integrating factor 

Compare this equation with the standard form of linear differential equation in the variable y is, $\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{y}}\mathbf{+}\mathit{P}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathit{y}\mathbf{=}\mathit{Q}\mathbf{\left(}\mathit{x}\mathbf{\right)}$

Now find the integrating factor$e^{\int P\left(x\right)dx}$.

$$\begin{gathered} e^{\int \left(\frac{1}{x}+1\right)dx}=e^{\ln x+x} \\ \Rightarrow e^{\ln x}·e^x \\ \Rightarrow x{e}^x \\ \left[\text{Since}{e}^{\ln x}=x\right] \end{gathered}$$

Thus, the integrating factor is$x{e}^x$

Determine the solution for given differential equation 

Multiply the equation (1) with an integrating factor Xe^x.

$$\begin{gathered} x{e}^x\left(y^{\text{'}}+\left(1+\frac{1}{x}\right)y\right)=x{e}^x\left(\frac{e^{-x}\sin 2x}{x}\right) \\ x{e}^x{y}^{\text{'}}+\left(x{e}^x+e^x\right)y=\sin 2x \\ \frac{d}{dx}\left(x{e}^{x}y\right)=\sin 2x \end{gathered}$$

Now integrate both side with respect to x.

$$\begin{gathered} \int \frac{d}{dx}\left(x{e}^{x}y\right)dx=\int \sin 2xdx \\ x{e}^{x}y=-\frac{\cos 2x}{2}+C \\ y=-\frac{1}{2x}{e}^{-x}\cos 2x+\frac{C{e}^{-x}}{x} \end{gathered}$$

Therefore, the solution of given differential equation is

$y=-\frac{1}{2x}{e}^{-x}\cos 2x+\frac{C{e}^{-x}}{x}$

The objective is to determine whether there are any transient terms in the general solution

Since $-\frac{1}{2x}{e}^{-x}\cos 2x\to 0and\frac{e^{-x}}{x}\to 0$ as$x\to \infty$

Since each term of general solution of differential equation is approaches to zero as x approaches to infinity. Therefore, the solution is transient.