General solution to 2nd order....

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Discussion Overview

The discussion revolves around finding a general solution to the second-order differential equation y" + 4y' + 3y = 4^(-t) using the method of undetermined coefficients. Participants explore the challenges associated with the non-standard exponential function on the right-hand side and share their attempts at finding particular solutions.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant expresses uncertainty about how to solve the equation due to the presence of 4^(-t) as the non-homogeneous term, noting a lack of examples with bases other than e.
  • Another participant suggests that if a proposed particular solution appears in the homogeneous solution, it should be multiplied by t to find a valid particular solution.
  • A participant attempts to differentiate their proposed particular solution y = Ct e^(-t) but reports that their calculations yield zero when substituted back into the original equation, raising doubts about their differentiation process.
  • Hints are provided regarding the transformation of 4^(-t) into an exponential form, suggesting that the particular solution could be of the form y_p(t) = A * 4^(-t).
  • One participant presents a potential particular solution in the form of yp(t) = 4^(-t)/(log^2(4) - 4(log(4)) + 3) and seeks confirmation of its correctness.
  • Another participant agrees with the proposed particular solution and acknowledges its correctness.
  • One participant revisits their earlier calculations and corrects their understanding of the right-hand side of the differential equation, indicating a misreading of the equation.

Areas of Agreement / Disagreement

Participants express differing views on the correct approach to finding the particular solution, with some proposing specific forms while others question their validity. The discussion remains unresolved regarding the best method to apply for this particular equation.

Contextual Notes

There are indications of confusion regarding the differentiation process and the interpretation of the right-hand side of the differential equation, which may affect the proposed solutions. Some assumptions about the form of the particular solution are also under discussion.

rayne1
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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).
 
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rayne said:
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.
 
Prove It said:
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?
 
Last edited:
Hint:

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

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

$$y_p(t)=A\cdot4^{-t}$$?
 
MarkFL said:
Hint:

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

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

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

Not sure if correct, but I got:

yp(t) = 4^(-t)/(log^2(4)-4(log(4))+3)
 
rayne said:
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)
 
rayne said:
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.
 

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