# Particular solution to nonhomogenous differential equation $y''-6y'+9y=9t+3$.

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 $e)$?

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.

• I have got $$y_p=t+1$$ check it. – Dr. Sonnhard Graubner Aug 3 '18 at 10:37
• Yes that's what I got as well, and results-list says that it is correct but d) and e) are correct as well which I don''t quite understand.. – JayFreddy Aug 3 '18 at 10:40
• Let $y=(C_1e^{3t}+C_2te^{3t})$ is general solution and $y=t$ is particulat, then $y_1=2e^{3t}+5te^{3t}+t$ is a solution. So if we separate $y_1$ and write $y_1=e^{3t}+5te^{3t}+(t+e^{3t})$, then $(t+e^{3t})$ is a particular solution as well??? – Nosrati Aug 3 '18 at 10:47
• How did you get that $y_h=2e^{3t} + t e^{3t}$ without any initial data? – Calvin Khor Aug 3 '18 at 10:48
• While it is true that $y_p = t+1$ is a Particular Solution (in the sense that the phrase presumably has a mathematical definition), its also true that the answers $a,d,e$ are particular solutions (i.e., they are examples of solutions) – Calvin Khor Aug 3 '18 at 10:51

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.
• oh okay, I thought that I had to use the $y(t) = 2e^{3t} + te^{3t}$ and not $y(t)=c_1 e^{3t}+c_2 te^{3t}$, it makes sence now, thx =) – JayFreddy Aug 3 '18 at 11:06
$$y=y_h+y_p=\bigl(c_1e^{3t}+c_2te^{3t}\bigr)+(t+1)$$
and the only compatible answers are $a$,$d$ and $e$.