# solve for second order linear differential equations [closed]

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

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

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

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• Have you noticed the repeated root ... if you show your working so far it should be simple to point out your mistake. – Mandelbrot Jun 28 '18 at 16:07
• 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 – Mandelbrot Jun 28 '18 at 16:15
• The good Dr Graubner has given you all you need .... – Mandelbrot Jun 28 '18 at 16:16

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 $y(t)= At^2e^{2t}$.

Hint: make for the particular solution the ansatz

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

• Thank you it makes sense now! – linda Jun 29 '18 at 2:37

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, $$\boxed{y(t)=e^{2t}(-3t^2+K_1t+K_2)}$$

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