Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Solve by undermined coefficients/variation of parameters: \[ y''-4y'+4y = \frac{e^{2x}}{x}\]

Differential Equations
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

\[ y_c''-4y_c'+4y_c=0\]\[ \lambda^2-4\lambda+4=0\]\[ \lambda=2, 2\]\[y_c = c_1e^{2x}+c_2xe^{2x}\] How can I guess the particular solution?
you can't do this one with undetermined cofficients. you would have to use variation of parameter.
to find the yp

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Why doesn't undetermined coefficients work in this case??
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?
for "x" general form of its solution of yp would be Ax+B
Is xe^(2x) a general form?
general for have unknown cofficients.. this one is not general form.. but this has a general for , it would be (Ax+B)e^(cx)
Then how do you define ''general''??
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
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''
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
\(x e^{2x}\) still qualifies as general
and e^(2x) / x on the other hand is not general?
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.
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
lol well i am glad it worked out for u :)
lol Thanks!! :)

Not the answer you are looking for?

Search for more explanations.

Ask your own question