How do I solve the DE $y''+6y'+8y = 2t+e^t$? The differential equation I am trying to solve is 
$$
\dfrac{d^2y}{dt^2} + 6\dfrac{dy}{dt} + 8y = 2t +e^t
$$
I know how to start off. I have done the $s^2 + 6s + 8 = 0$ to get $s = -4$ and $s = -2$ and have the 
$$
y_p(t) = k_1e^{-4t} + k_2e^{-2t}
$$
What I am having a problem with is gettting $y_h(t)$ which will be dependent on the $2t + e^t$. I know that the solution will be in the form $c_1 + c_2t + c_3e^t$, I am just not sure how to solve for those constants
Thank you very much for any help you can give me.
 A: If you accept $Dy=y'$ so the OE can be rewritten as $$(D^2+6D+8)y=2t+\text{e}^t$$ Moreover, you can see that $D(2t)=2$ and $D(2)=0$ so $D^2(2t)=0$. If fact $D^2$ annihilates the function $2t$. With the same approach, you'll find out that $(D-1)\text{e}^t=0$ (Try to show that!). Let's get to gather what we have had here. 


*

*$(D^2+6D+8)y=2t+\text{e}^t$

*$D^2(2t)=0$

*$(D-1)\text{e}^t=0$
From 2 and 3, we get $D^2(D-1)(2t+\text{e}^t)=0$  and by considering 1, we have $$D^2(D-1)(D^2+6D+8)y=D^2(D-1)(2t+\text{e}^t)=0$$ It means that the differential polynomial $$P(D)=D^2(D-1)(D^2+6D+8)=D^2(D-1)(D+2)(D+4)$$ annihilates the function $y$.

Now, let’s to forget the point we stand on and let someone gave you a $P(D)$ as $$P(D)=D^2(D-1)(D+2)(D+4)$$ and he said you that this polynomial annihilates an unknown function $y$. In fact, he is asking you to guess what is the probable function $y$. I say:


*

*As we have a $D^2$ term so there are $C_1+C_2t$ as two terms in $y$.

*As we have a $D-1$ term in $P(D)$, so there is a term like $C_3\text{e}^t$ in $y$.

*As we have a $D+2$ term and a $D+4$ term in $P(D)$, so there are two terms like $C_4\text{e}^{-2t}$ and $C_5\text{e}^{-4t}$ in $y$.
Overall, the desired function $y$ has the form $$y=C_1+C_2t+C_3\text{e}^t+C_4\text{e}^{-2t}+C_5\text{e}^{-4t}$$ Your $y_h$ has the form $y_h=C_1+C_2t+C_3\text{e}^t$. Substitute it to your OE and find the proper constants $C_1,C_2,C_3$.
