General solution to 2nd order....

[rayne1]

I'm supposed to use undetermined coefficients to find a general solution to:
y" + 4y' +3y =4^(-t)

I can't find an example online where f(t) is equal to an exponential function that does not have e as the base, so I have no idea how to solve it.

So far, I found the general solution to the homogeneous equation which is:
y(t) = Ae^(-1t) + Be^(-3t).

 

[Prove It]

I'm supposed to use undetermined coefficients to find a general solution to:
y" + 4y' +3y =4^(-t)

I can't find an example online where f(t) is equal to an exponential function that does not have e as the base, so I have no idea how to solve it.

So far, I found the general solution to the homogeneous equation which is:
y(t) = Ae^(-1t) + Be^(-3t).

When what you try for a particular solution appears in your homogeneous solution, multiply it by t. If that appears in it, multiply that by t, etc...

Here $\displaystyle \begin{align*} y = C\,t\,\mathrm{e}^{-t} \end{align*}$ should work.

 

[rayne1]

When what you try for a particular solution appears in your homogeneous solution, multiply it by t. If that appears in it, multiply that by t, etc...

Here $\displaystyle \begin{align*} y = C\,t\,\mathrm{e}^{-t} \end{align*}$ should work.

I tried and it didn't work. Don't know if I made a mistake in differentiating but this is what I did:
y = Cte^(-t)
y' = Ce^(-t)-Cte^(-t)
y'' = -Ce^(-t)-Ce^(-t)-Cte^(-t)

When I plugged it back into the original equation, I got 0.

Edit: Oh, Is y" supposed to equal -Ce^(-t)-Ce^(-t)+Cte^(-t), instead?

 

[MarkFL]

Hint:

\(\displaystyle 4^{-t}=e^{-\ln(4)t}\)

So, can you show that the particular solution must be of the form:

\(\displaystyle y_p(t)=A\cdot4^{-t}\)?

 

[rayne1]

Hint:

\(\displaystyle 4^{-t}=e^{-\ln(4)t}\)

So, can you show that the particular solution must be of the form:

\(\displaystyle y_p(t)=A\cdot4^{-t}\)?

Not sure if correct, but I got:

yp(t) = 4^(-t)/(log^2(4)-4(log(4))+3)

 

[MarkFL]

Not sure if correct, but I got:

yp(t) = 4^(-t)/(log^2(4)-4(log(4))+3)

Yes, that agrees with W|A. Good job! (Yes)

 

[Prove It]

I tried and it didn't work. Don't know if I made a mistake in differentiating but this is what I did:
y = Cte^(-t)
y' = Ce^(-t)-Cte^(-t)
y'' = -Ce^(-t)-Ce^(-t)-Cte^(-t)

When I plugged it back into the original equation, I got 0.

Edit: Oh, Is y" supposed to equal -Ce^(-t)-Ce^(-t)+Cte^(-t), instead?

Yes it should be $\displaystyle \begin{align*} y'' = -C\,\mathrm{e}^{-t} - C\,\mathrm{e}^{-t} + C\,t\,\mathrm{e}^{-t} = C\,t\,\mathrm{e}^{-t} - 2C\,\mathrm{e}^{-t} \end{align*}$...

- - - Updated - - -

I apologise, for some reason I read the RHS of your DE as $\displaystyle \begin{align*} 4\mathrm{e}^{-t} \end{align*}$, not $\displaystyle \begin{align*} 4^{-t} \end{align*}$. Please disregard my other posts in this thread.