Question

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

Answer 1

I love the method of undetermined guessing.

Answer 2

\[y_{c}'-3y_c = 0\]\[y_{c}=ce^{3x}\]\[\text{Try } y_{p1}=Ce^{2x}\]\[2Ce^{2x}-3Ce^{2x} = xe^{2x}\]\[-Ce^{2x} = xe^{2x}\]\[C = -x\]
\[y_{p2}=D\]\[0 - 3D = 6\]\[D=-2\]
\[y= ce^{3x}-xe^{2x}-2\]?!
No!!~~~

Answer 3

''Educated guess'' lol

Answer 4

2Ce2x−3Ce2x=xe2x

Answer 5

the particular solution is \[\huge y_p=Be^{2x}+Cxe^{2x}+D\]

Answer 6

going to need something like e^2x (A+Bx)

Answer 7

Hmm... Let me try again!!!

Answer 8

\[y_{p1} = e^{2x}(A+Bx)\]\[2e^{2x}(A+Bx)+Be^{2x}-3e^{2x}(A+Bx)=xe^{2x}\]\[Be^{2x}-e^{2x}(A+Bx)=xe^{2x}\]\[e^{2x}(A-B+Bx)=-xe^{2x}\]\[A=B=-1\]
\[y_{p1} = e^{2x}(-1-x)\]
\[y_{p2} = -2\]
\[y= ce^{3x} -e^{2x}-xe^{2x}-2\]
It works!!! o_O

Answer 9

nice work.

Answer 10

How did you get this: \(y_p=Be^{2x}+Cxe^{2x}+D\) ??

Answer 11

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 \[\huge \left( C_0 + C_1x + C_2 x^2 + ... + C_n x^n\right) e^{2x}\]

Answer 12

Does the coefficient of x in e^{nx} matter?

Answer 13

yes. that determines the particular solution

Answer 14

Hmm... In what way?

Answer 15

the polynomial factor of e^{ax}, that is
\[\large c_0 + c_1x + c_2x^2 + ... c_nx^n\]

Answer 16

Why is particular solution for this one is e^(2x) (A+Bx)
But it doesn't work for y' - 3y = xe^(3x) + 4???

Answer 17

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 \[\large (Bx + Cx^2)e^{3x}\]

Answer 18

That's complicated :(