A community for students.
Here's the question you clicked on:
 0 viewing
RolyPoly
 3 years ago
Obtain a general solution using the method of undetermined coefficients.
\[y'3y=xe^{2x}+6\]
RolyPoly
 3 years ago
Obtain a general solution using the method of undetermined coefficients. \[y'3y=xe^{2x}+6\]

This Question is Closed

Algebraic!
 3 years ago
Best ResponseYou've already chosen the best response.0I love the method of undetermined guessing.

RolyPoly
 3 years ago
Best ResponseYou've already chosen the best response.1\[y_{c}'3y_c = 0\]\[y_{c}=ce^{3x}\]\[\text{Try } y_{p1}=Ce^{2x}\]\[2Ce^{2x}3Ce^{2x} = xe^{2x}\]\[Ce^{2x} = xe^{2x}\]\[C = x\] \[y_{p2}=D\]\[0  3D = 6\]\[D=2\] \[y= ce^{3x}xe^{2x}2\]?! No!!~~~

RolyPoly
 3 years ago
Best ResponseYou've already chosen the best response.1''Educated guess'' lol

sirm3d
 3 years ago
Best ResponseYou've already chosen the best response.3the particular solution is \[\huge y_p=Be^{2x}+Cxe^{2x}+D\]

Algebraic!
 3 years ago
Best ResponseYou've already chosen the best response.0going to need something like e^2x (A+Bx)

RolyPoly
 3 years ago
Best ResponseYou've already chosen the best response.1Hmm... Let me try again!!!

RolyPoly
 3 years ago
Best ResponseYou've already chosen the best response.1\[y_{p1} = e^{2x}(A+Bx)\]\[2e^{2x}(A+Bx)+Be^{2x}3e^{2x}(A+Bx)=xe^{2x}\]\[Be^{2x}e^{2x}(A+Bx)=xe^{2x}\]\[e^{2x}(AB+Bx)=xe^{2x}\]\[A=B=1\] \[y_{p1} = e^{2x}(1x)\] \[y_{p2} = 2\] \[y= ce^{3x} e^{2x}xe^{2x}2\] It works!!! o_O

RolyPoly
 3 years ago
Best ResponseYou've already chosen the best response.1How did you get this: \(y_p=Be^{2x}+Cxe^{2x}+D\) ??

sirm3d
 3 years ago
Best ResponseYou've already chosen the best response.3if the right hand side is \[\large e^{2x}\] the corresponding particular solution is \[\large Ce^{2x}\] but if the RHS is some expression \[\large x^{n}e^{2x}\] then the particular solution is \[\huge \left( C_0 + C_1x + C_2 x^2 + ... + C_n x^n\right) e^{2x}\]

RolyPoly
 3 years ago
Best ResponseYou've already chosen the best response.1Does the coefficient of x in e^{nx} matter?

sirm3d
 3 years ago
Best ResponseYou've already chosen the best response.3yes. that determines the particular solution

sirm3d
 3 years ago
Best ResponseYou've already chosen the best response.3the polynomial factor of e^{ax}, that is \[\large c_0 + c_1x + c_2x^2 + ... c_nx^n\]

RolyPoly
 3 years ago
Best ResponseYou've already chosen the best response.1Why is particular solution for this one is e^(2x) (A+Bx) But it doesn't work for y'  3y = xe^(3x) + 4???

sirm3d
 3 years ago
Best ResponseYou've already chosen the best response.3it will not work since the complementary solution is \[\large c_1e^{3x}\]. the constant coefficient is taken by the complementary solution, therefore the polynomial factor of the particular solution "begins with" \[\large Bx\]. Thus the particular solution is \[\large (Bx + Cx^2)e^{3x}\]
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.