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RolyPoly

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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  1. Algebraic!
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    I love the method of undetermined guessing.

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

    • one year ago
  3. RolyPoly
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    ''Educated guess'' lol

    • one year ago
  4. Algebraic!
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    2Ce2x−3Ce2x=xe2x

    • one year ago
  5. sirm3d
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    the particular solution is \[\huge y_p=Be^{2x}+Cxe^{2x}+D\]

    • one year ago
  6. Algebraic!
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    going to need something like e^2x (A+Bx)

    • one year ago
  7. RolyPoly
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    Hmm... Let me try again!!!

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

    • one year ago
  9. sirm3d
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    nice work.

    • one year ago
  10. RolyPoly
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    How did you get this: \(y_p=Be^{2x}+Cxe^{2x}+D\) ??

    • one year ago
  11. sirm3d
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    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
  12. RolyPoly
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    Does the coefficient of x in e^{nx} matter?

    • one year ago
  13. sirm3d
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    yes. that determines the particular solution

    • one year ago
  14. RolyPoly
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    Hmm... In what way?

    • one year ago
  15. sirm3d
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    the polynomial factor of e^{ax}, that is \[\large c_0 + c_1x + c_2x^2 + ... c_nx^n\]

    • one year ago
  16. RolyPoly
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    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
  17. sirm3d
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    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
  18. RolyPoly
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    That's complicated :(

    • one year ago
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