find the solutions of $y^{\prime \prime}-4 y^{\prime}+3 y=8 e^{-x}+9$ s.t $\lim _{x \rightarrow \infty} e^{-x} y(x)=7$ I have the ODE $$y^{\prime \prime}-4 y^{\prime}+3 y=8 e^{-x}+9$$
I am asked to find a solution such that: $$\lim _{x \rightarrow \infty} e^{-x} y(x)=7$$
This question feels a liitle bit tricky, how should I approach for such that question?
should I first solve it like a regular ODE without boundary condition?
 A: We can solve the homogeneous case using the characteristic equation method and find
$$y_H(x)=c_1 e^{x} + c_2 e^{3x}$$
Now we guess a particular solution. This can be done more systematically using variation of parameters, but in this case guessing is rather easy. We guess
$$y_P(x)=Ae^{-x}+B$$
Then
$${y_P}^{\prime}(x)=-A e^{-x} \text{and } {y_P}^{\prime \prime}(x)=A e^{-x}$$
Substituting into our ODE,
$$A e^{-x} +4Ae^{-x} + 3(Ae^{-x}+B)=8e^{-x}+9$$
So then
$$(A+4A+3A)e^{-x}+3B=8e^{-x}+9$$
Clearly $A=1, B=3$.
Then, $y(x)=y_H(x)+y_P(x)$, so our solution is
$$y(x)= c_1 e^{x} + c_2 e^{3x} + e^{-x}+3$$
So,
$$\lim_{x\to\infty}e^{-x}y(x)=\lim_{x\to\infty}(c_1+c_2e^{2x}+e^{-2x}+3e^{-x})$$
In order for the limit to be finite, $c_2=0$. Then,
$$\lim_{x\to\infty}(c_1+e^{-2x}+3e^{-x})=c_1$$
Therefore $c_1=7$. Therefore our solution is
$$y(x)=7e^x + e^{-x} +3.$$
A: Consider $u(x)=e^{-x}y(x)$. Because this has an asymptotic limit for $x\to\infty$, $u'(x)$ has to converge to zero there.
Then $u'(x)=e^{-x}(y'(x)-y(x))$, $u''(x)=e^{-x}(y''(x)-2y'(x)+y(x))$ and thus
$$
u''(x)-2u'(x)=e^{-x}(y''(x)-4y'(x)+3y(x))=8e^{-2x}+9e^{-x}
$$
This can be integrated once to
$$
u'(x)-2u(x)=-4e^{-2x}-9e^{-x}+C
$$
and because of the limit, $C=-14$. The next integrating factor is $e^{-2x}$ so that
$$
(e^{-2x}u(x))'=-4e^{-4x}-9e^{-3x}-14e^{-2x}
\\
u(x)=e^{-2x}+3e^{-x}+7
$$
with no further integration constant, as the corresponding term would diverge at infinity.
