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anonymous
 3 years ago
Obtain a general solution using the method of undetermined coefficients.
\[y'3y=xe^{2x}+6\]
anonymous
 3 years ago
Obtain a general solution using the method of undetermined coefficients. \[y'3y=xe^{2x}+6\]

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0''Educated guess'' lol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the particular solution is \[\huge y_p=Be^{2x}+Cxe^{2x}+D\]

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hmm... Let me try again!!!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[y_{p1} = e^{2x}(A+Bx)\]\[2e^{2x}(A+Bx)+Be^{2x}3e^{2x}(A+Bx)=xe^{2x}\]\[Be^{2x}e^{2x}(A+Bx)=xe^{2x}\]\[e^{2x}(AB+Bx)=xe^{2x}\]\[A=B=1\] \[y_{p1} = e^{2x}(1x)\] \[y_{p2} = 2\] \[y= ce^{3x} e^{2x}xe^{2x}2\] It works!!! o_O

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0How did you get this: \(y_p=Be^{2x}+Cxe^{2x}+D\) ??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if 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 \[\huge \left( C_0 + C_1x + C_2 x^2 + ... + C_n x^n\right) e^{2x}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Does the coefficient of x in e^{nx} matter?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes. that determines the particular solution

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the polynomial factor of e^{ax}, that is \[\large c_0 + c_1x + c_2x^2 + ... c_nx^n\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Why is particular solution for this one is e^(2x) (A+Bx) But it doesn't work for y'  3y = xe^(3x) + 4???

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it 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 \[\large (Bx + Cx^2)e^{3x}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That's complicated :(
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