-2
$\begingroup$

I got the sum of A is 0? There is no solution to this? Can someone please help. Thanks!

$$y''-4y'+4y=-6e^{2t}$$

$\endgroup$

closed as off-topic by user99914, user296602, Batominovski, Jyrki Lahtonen, Vlad Jun 28 '18 at 21:28

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, Community, Jyrki Lahtonen, Vlad
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Have you noticed the repeated root ... if you show your working so far it should be simple to point out your mistake. $\endgroup$ – Mandelbrot Jun 28 '18 at 16:07
  • $\begingroup$ What did you get for your general solution to the homogeneous case and what are you going to use for your trial solution ... just be careful you don't choose $y=Ae^2t$ because it should be in your GS $\endgroup$ – Mandelbrot Jun 28 '18 at 16:15
  • $\begingroup$ The good Dr Graubner has given you all you need .... $\endgroup$ – Mandelbrot Jun 28 '18 at 16:16
0
$\begingroup$

I have no idea what you mean by "sum of A". There is no "A" in the problem. Do you mean that you are looking for a "particular solution" of the form $Ae^{2t}$? If so you should NOT be getting A= 0- you should not be getting anything at all for A! Setting $y= Ae^{2t}$, $y'= 2Ae^{2t}$ and $y''= 4Ae^{2t}$ so that $y''- 4y+ 4= 4Ae^{2t}- 8Ae^{2t}+ 4Ae^{2t}= 0$ no matter what A is.

The "double root" that Bruce refers to is the double root of the characteristic equation: the associated homogeneous differential equation, y''- 4y'+ 4y= 0, has characteristic equation $r^2- 4r+ 4= (r- 2)^2= 0$ which has the double root r= 2. The general solution to the associated homogeneous equation is $y(t)= C_1e^{2t}+ C_2te^{2t}$. You need to look for a specific solution of the form $y(t)= At^2e^{2t}$.

$\endgroup$
1
$\begingroup$

Hint: make for the particular solution the ansatz

$$y_p=Ae^{2t}t^2$$

$\endgroup$
  • $\begingroup$ Thank you it makes sense now! $\endgroup$ – linda Jun 29 '18 at 2:37
0
$\begingroup$

Try this as particular solution $$y_p=At^2e^{2t}$$


Another solution $$y''-4y'+4y=-6e^{2t}$$ Rewriting the differential equation $$(y''-2y')-2(y'-2y)=-6e^{2t}$$ $$(y'e^{-2t})'-2(ye^{-2t})'=-6$$ Integrate $$y'e^{-2t}-2ye^{-2t}=-6t+K_1$$ $$(ye^{-2t})'=-6t+K_1$$ Integrate again $$ye^{-2t}=-3t^2+K_1t+K_2$$ Finally we get, $$\boxed{y(t)=e^{2t}(-3t^2+K_1t+K_2)}$$

$\endgroup$
  • 1
    $\begingroup$ thank you so much! $\endgroup$ – linda Jun 29 '18 at 2:37
  • $\begingroup$ yw @linda .............. $\endgroup$ – Isham Jun 29 '18 at 4:48

Not the answer you're looking for? Browse other questions tagged or ask your own question.