Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
RolyPoly
Group Title
Obtain a general solution using the method of undetermined coefficients.
\[y'3y=xe^{2x}+6\]
 2 years ago
 2 years ago
RolyPoly Group Title
Obtain a general solution using the method of undetermined coefficients. \[y'3y=xe^{2x}+6\]
 2 years ago
 2 years ago

This Question is Closed

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

RolyPoly Group TitleBest 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!!~~~
 2 years ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
''Educated guess'' lol
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.0
2Ce2x−3Ce2x=xe2x
 2 years ago

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

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

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
Hmm... Let me try again!!!
 2 years ago

RolyPoly Group TitleBest 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
 2 years ago

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

sirm3d Group TitleBest ResponseYou've already chosen the best response.3
if 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 \[\huge \left( C_0 + C_1x + C_2 x^2 + ... + C_n x^n\right) e^{2x}\]
 2 years ago

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

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

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
Hmm... In what way?
 2 years ago

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

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

sirm3d Group TitleBest ResponseYou've already chosen the best response.3
it 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 \[\large (Bx + Cx^2)e^{3x}\]
 2 years ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
That's complicated :(
 2 years ago
See more questions >>>
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.