Question

Find the general solution of the given differential equation

(x+!) dy/dx + (x+2)y = 2xe^-x

y = ______

Determine whether there are any transient terms in the general solution.

Answer 1

Divide throughout by $(x+1)$ :

$\frac{dy}{dx} + \frac{(x+2)}{(x+1)} y= \frac{2xe^{-x }}{(x + 1)}$

$IF = e^{\int \frac{(x+2)}{(x+1)}dx}$

$IF = e^{\int 1+ \frac{1}{(x+1)}dx}$

$IF = e^{x+ln|x+1|}$

$IF = (x + 1)e^{x}$

Multiply throughout by the Integrating factor :

$(x + 1)e^{x} * [\frac{dy}{dx} + \frac{(x+2)}{(x+1)} y = \frac{2xe^{-x} }{ (x + 1)}]$

$\frac{d}{dx}(y (x + 1)e^{x}) = (x + 1)e^{x} \frac{2xe^{-x}}{(x + 1)}$

$\frac{d}{dx}(y (x + 1)e^{x}) =2x$

Integrating :

$y(x + 1)e^{x} = x^2 + C$

$y = \frac{x^2 + C}{(x+1)e^x}$

Clearly, since we have the $x + 1$ in denominator, x cannot be equal to -1

This splits number line into intervals $(- \infin , -1)$ and $(-1 , \infin)$

out of which $(-1 , \infin)$ is LARGER

So, $(-1 , \infin)$ $\implies$ ANSWER

Both terms, $\frac{x^2}{(x+1)e^x}$ and $\frac{C}{(x+1)e^x}$ are transient because as $x\to \infin$ , both of these terms tend to 0.

Answer: $\frac{x^2}{(x+1)e^x},\frac{C}{(x+1)e^x}$ .