Question

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

$\mathbf{ydx}\mathbf{-}\mathbf{(}{\mathbf{ye}}^{\mathbf{y}}\mathbf{-}\mathbf{2}\mathbf{x}\mathbf{)}\mathbf{dy}\mathbf{=}\mathbf{0}\mathbf{.}$

Short answer

The general solution of the given differential equation is $x=e^y-\frac{2}{y}{e}^y+\frac{2}{y^2}{e}^y+\frac{c}{y^2}$and

there are transient terms in the general solution.

Step-by-step solution

Rewriting the given equation in the standard form

Consider the given differential equation,

$$\begin{gathered} ydx-\left(y{e}^y-2x\right)dy=0 \\ ydx=\left(y{e}^y-2x\right)dy \\ \frac{dx}{dy}=\frac{y{e}^y-2x}{y} \\ \frac{dx}{dy}+\frac{2}{y}x=e^y........\left(1\right) \end{gathered}$$

The standard form of the differential equation is,

$\frac{dx}{dy}+P\left(y\right)x=f\left(y\right)$

Here, x is dependent variable y is independent variable

Determine the integration factor

Now, equation (1) is in the standard form with

$P\left(y\right)=\frac{2}{y}$

So, integrating factor is,

$$\begin{gathered} e^{\int u\left(y\right)dy}=e^{\int \frac{2}{y}dy} \\ \Rightarrow e^{21dy} \\ \Rightarrow y^2 \end{gathered}$$

Determine the general solution for the given differential equation

Multiplying (1) with integrating factor as follows: which are continuous on $\left(0,\infty \right)$.

$$\begin{gathered} y^2\frac{dx}{dy}+2yx=y^2{e}^y,\text{which is same as} \\ \frac{d}{dy}\left[y^{2}x\right]=y^{2}e \end{gathered}$$

Integrating the above equation on both sides yields,

Use integration by parts method to find the integration

$$\begin{gathered} I=\int y^2{e}^{y}dy \\ I=y^2\int e^{y}dy-\int \left(\frac{d\left(y^2\right)}{dy}\int e^{y}dydy\right. \\ I=y^2{e}^y-\int \left(2y{e}^y\right)dy \\ I=y^2{e}^y-2\left[y\int e^{y}dy-\int \left(\frac{dy}{dy}\int e^{y}dy\right)dy\right] \\ I=y^2{e}^y-2\left[y{e}^y-\int e^{y}dy\right] \\ I=y^2{e}^y-2y{e}^y+e^y+c\int y^2{e}^{y}dy=e^y\left(y^2-2y+2\right)+c \end{gathered}$$

Substitute the value $I=\int y^2{e}^{y}dy$in the equation (1) obtain,

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

Determine there are any transient terms in the general solution.

The objective is to find the general solution of this differential equation and give the largest interval at which the general solution is defined.

Identify whether there are any transient terms in the general solution or not.

As$y\to \infty ;\frac{c}{y^2}\to 0$

Hence $\frac{c}{y^2}$is a transient.