solving $y'+5y=12e^{-t}$ using undetermined coeffecients method From "Circuit Analysis Demystified", David McMahon, 2008, Chapter 6, page 131, Quiz question 4.
Solve DE using undetermined coefficients method: 
$$y'+5y=12e^{-t}$$
Step 1: find homogeneous solution
$$y'+5y=0$$
$$r + 5 = 0$$
$$r = -5$$
$$y_h = Ae^{-5t}$$
Step 2: find particular solution
$$y_p = Bte^{-t}$$
$$y_p' = At\frac{d}{dt}\Big(e^{-t}\Big) + e^{-t}\frac{d}{dt}\Big(At\Big)$$
$$y_p' = -Ate^{-t} + Ae^{-t}$$
substituting into in-homogenous DE:
$$Ae^{-t}(1-t) + 5A~t~e^{-t} = 12 e^{-t}$$
$$A(1-t) +5At=12$$
$$A-At + 5At=12$$
$$A(1+4t)=12$$
$$A = \frac{12}{1+4t}$$
Therefore particular solution is:
$$v_p = \frac{1}{1+4t}t e^{-t}$$
And my Total solution is:
$$y = y_h + y_p$$
$$y = Ae^{-5t} + \frac{1}{1+4t}t e^{-t}$$
Questions:
The answer key in the back of the text book says the solution to the DE is:
$$y = 3e^{-5t} (e^{4t}-1)$$
How did they get that using indeterminate coefficients method?
Also, is it ok to use $y_p = Ae^{-t}$ as solution for step 2?  I was told by another textbook not to use that solution again because its the same as the homogeneous solution...but to modify it as $y_p=Ate^{-t}$, but maybe this is a special case where you can do that again?
 A: Trying $y_p(t)=Ae^{-t}$. $y_p'+5y_p=-Ae^{-t}+5Ae^{-t}=4Ae^{-t}$. Setting it equal to $12e^{-t}$, we get that $A=3$. Note that $y_p$ is not a solution of the homogeneous equation, even if it is an exponential. If you had, for example, $y'+5y=e^{-5t}$, you'd then try a solution of the form $Ate^{-5t}$, where $A$ is a constant. The coefficient in front of the $t$ matters here!
A: Your homogenous solution is $y = Ae^{-5t}$ whereas the RHS of the original equation is $12e^{-t}$. Notice that the powers of $-5t$ and $-t$ are not the same and so you should simply use $Be^{-t}$.
A: First of all, I agree that $y_h(t)=Ae^{-5t}$. However, if you look at the right hand side, the exponential is not of the same power as -5. Hence, $y_p$ should be of the form $y_p=Be^{-t}$. Indeed, plugging in, we have
$$
y_p' + 5y_p = Be^{-t}(-1+5)=12e^{-t}
$$
so B = 3. The general solution is then
$$
y=Ae^{-5t} + 3e^{-t}.
$$
For this to match up with the back of the book, my guess is that there's an initial condition supplied (i.e. $y(0)=0$), which makes $A=-3.$ Hence,
$$
y=3e^{-5t}(e^{4t}-1).
$$
A: Adding to the good answers that pointed the mistake of the particular solution:
$$Ae^{-t}(1-t) + 5A~t~e^{-t} = 12 e^{-t}$$
Note that this must be true for all t..
So that we have:
$$A(1-t) + 5A~t~= 12 $$
$$ t(4A)+(A- 12)=0 $$
$$\implies 4A=0 \text {, and } A-12=0$$
And it dosen't exist an A that fill both conditions. Your deduction about A as a function f t is not correct $(A=A(t)$. A is a constant. Because if A was a function of $t$ then your derivative is wrong in the first place. In this line $A$ is a constant. 
$$y_p' = At\frac{d}{dt}\Big(e^{-t}\Big) + e^{-t}\frac{d}{dt}\Big(At\Big)$$
But you end with $A=A(t)$. 
The particular solution should be:$$y_p=Ae^{-t}$$
