## RolyPoly 2 years ago Solve by undermined coefficients/variation of parameters: $y''-4y'+4y = \frac{e^{2x}}{x}$

1. RolyPoly

$y_c''-4y_c'+4y_c=0$$\lambda^2-4\lambda+4=0$$\lambda=2, 2$$y_c = c_1e^{2x}+c_2xe^{2x}$ How can I guess the particular solution?

2. nubeer

you can't do this one with undetermined cofficients. you would have to use variation of parameter.

3. nubeer

to find the yp

4. RolyPoly

Why doesn't undetermined coefficients work in this case??

5. nubeer

well in undetermined cofficients there are few general forms of the solution already told.. i mean like if f(x) is e^x = Ae^ax.. and a few more like these.. know of it?

6. nubeer

for "x" general form of its solution of yp would be Ax+B

7. RolyPoly

Is xe^(2x) a general form?

8. nubeer

general for have unknown cofficients.. this one is not general form.. but this has a general for , it would be (Ax+B)e^(cx)

9. RolyPoly

Then how do you define ''general''??

10. nubeer

like for equation of a line y=mx+c.. this is general form. u ahve to find x and m.. same way in here yp = (Ax+B)e^(cx) is general form and we have to find A,B,c. th

11. RolyPoly

Some typical examples are constant - constant sin / cos - Asinx + Bcosx e^(ax) = Ce^(ax) x^n = polynomial of x But xe^(2x) is not that ''general''

12. nubeer

in exams we never get the straight one .. in here we have to first see if x has general for or not.. then we would see if e^2x has general form or nto.. if they both would have just miltiply them

13. sirm3d

$$x e^{2x}$$ still qualifies as general

14. RolyPoly

and e^(2x) / x on the other hand is not general?

15. nubeer

in my opinion, No. because i think that x in the denominator part is the trouble here and we dont have a general form for that part. so i think variation of parameter should do the trick.

16. RolyPoly

Variation of parameters: $y'' - 4y' + 4y = \frac{e^{2x}}{x}$$y_c'' - 4y_c' + 4y_c =0$$y_c = c_1e^{2x}+c_2xe^{2x}$ $u_1'e^{2x} + u_2' xe^{2x}=0$$2u_1'e^{2x} + u_2' (2xe^{2x}+e^{2x})=\frac{e^{2x}}{x}$ $u_2'e^{2x}=\frac{e^{2x}}{x}$$u_2 = lnx$ $u_1' = -\frac{\frac{1}{x}xe^{2x}}{e^2x}$$u_1 = -x$ $y=c_1e^{2x}+c_2xe^{2x} +xlnxe^{2x} - xe^{2x}$ I can't believe that I got the solution! :O

17. nubeer

lol well i am glad it worked out for u :)

18. RolyPoly

lol Thanks!! :)