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RolyPoly
Group Title
Obtain a general solution using the method of undetermined coefficients.
\[y'3y=xe^{2x}+6\]
 one year ago
 one year ago
RolyPoly Group Title
Obtain a general solution using the method of undetermined coefficients. \[y'3y=xe^{2x}+6\]
 one year ago
 one year ago

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

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

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

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

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

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

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
How did you get this: \(y_p=Be^{2x}+Cxe^{2x}+D\) ??
 one year 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 \[\huge \left( C_0 + C_1x + C_2 x^2 + ... + C_n x^n\right) e^{2x}\]
 one year ago

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

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

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

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

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
That's complicated :(
 one year ago
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