I continue my quest of finishing a problem where I
previously calculated the unique solution
to a second order homogeneous differential equation and got the right answer in the end.
Now though, I have to Determine which of the following function expressions is a (particular) solution of the non-homogeneous differential equation
$$y''-6y'+9y=9t+3 \tag{1}\label{original}$$
with the following possible answers:
$$ \begin{array}{l|l}
a) t+1 & d)e^{3t}+t+1 \\
b) t^2+\frac{t}{2}-1 & e)-2te^{3t}+t+1 \\
c) 9t+\frac{17}{3} & f)e^{3t}-te^{3t} \\
\end{array} $$
I understand that the general solution, $y$, equals $y_h+y_p$ where $y_h$ is the general solution to the homogeneous equation $\bigl(y_h(t)=2e^{3t}+te^{3t} \bigr)$ and $y_p$ is any particular solution to the non-homogeneous equation. And if I follow the theory of my text book and Sal Khan from Khan Academy then I have to make a shrewd guess of what $y_p$ might be. So, since the differential equation equals a polynomial my guess of $y_p$ will be the polynomial:
$y_p=At+B \tag{2}\label{y_p}$ which gives us,
$y_p'=A$ and
$y_p''=0$
I then substitute these back into the original equation (\ref{original}) and get:
$$1\cdot (0) -6 \cdot (A) + 9\cdot(At+B)=9t+3\Rightarrow 9At-6A+9B=9t+3$$
We know that whatever the t-coefficients add up to be on the left-hand side they should be equal to the right-hand side, same goes for the constants. Let's calculate and solve for A and B:
$9A=9 \Rightarrow A=1$
$-6 \cdot 1 + 9B=3 \Rightarrow 9B = 3 + 6 \Rightarrow B=1$
We insert A and B into (\ref{y_p})
$y_p=t+1$
$y=y_h+y_p=\bigl(2e^{3t}+te^{3t}\bigr)+(t+1) \tag{3}\label{y}$
Final question: How would $y$ look when added together and why does the results-list say the correct answers are $a), d)$ and $e)$?
Thanks in advance =)
Edit:
I think that the professor wants the particular solution as well as the solution of the non-homogeneous D.E. because of the way he puts parentheses around the word particular, and that's why $a)$ is correct, but then I just need to understand how $d)$ and $e)$ are results as well.
Best Answer
Let $y=C_1e^{3t}+C_2te^{3t}$ is the general solution of the equation and $y=t+1$ is the particular solution, then $$y_0=C_1e^{3t}+C_2te^{3t}+t+1$$ is a particular solution for every specific numbers $C_1$ and $C_2$. For instance $$y_0=6e^{3t}+5te^{3t}+(t+1)=5e^{3t}+5te^{3t}+(t+1+e^{3t})$$ is a particular solution as well.