Question

If y(t)=sin(5t) is the solution of the ordinary differential equation: y''+ y' -6y =f(t) where f(t) is a function of t, which of the following must also be a solution of this d.e?
a. y(t)=2cos(5t)
b. y(t)=5e^2t +2sin(5t)
c. y(t)=-3e^-3t +2e^2t
d. y(t)=2sin(5t)-e^-3t
e. y(t)=-e^2t +2cos(5t)
f. y(t)=2sin(5t) -2e^(1-3t)
g. y(t)=10e^-2t -e^3t +2sin(5t)
h. y(t)=e^-3t +sin(5t)

Any help would be greatly appreciated because I'm really confused by this question.

Answer 1

Remember that solutions of differential equations are functions.
You are given a solution. So you know one y. If you differentiate it twice, you also know y' and y''. Plug these values of y, y'and y'' in the eq to see what is going on...

Answer 2

can you use variation of parameters ? to solve for the second function

Answer 3

I'll give variation of parameters a shot! thanks!

Answer 4

When I use variation of parameters, I get the following results. I think I'm doing something wrong.

The auxiliary equation is: \[m ^{2}+m-6\]=0 which factors to (m-3)(m+2)=0.
m=-2, or 3.

Then \[u_{1}^{'} y _{1} + u _{2}^{'}y _{2}=0\]
\[u _{1}^{'} e ^{-2t}+u _{2}^{'} e ^{3t}=0\]
\[u _{1}^{'}=-u _{2}^{'} e ^{5t}\]

Using that, the second equation becomes
\[u_{1}^{'} y _{1}^{'} + u _{2}^{'}y _{2}^{'}=0\]
\[-3e ^{8t}u _{2}^{'}-2e ^{-2t}u _{2}^{'}=f(t)\]
\[u _{2}^{'}(-3e ^{8t}-2e ^{-2t})=f(t)\]
\[u _{2}^{'}=\frac{ f(t) }{ -3e ^{8t}-2e ^{-2t} }\]
Then you are supposed to integrate both sides to get the value of \[u _{2}\] but I don't know how I would integrate due to the f(t) value. Any suggestions?