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anonymous

  • 5 years ago

Solve the ODE y''''-3y'''+4''=2(x-1)e^x

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  1. anonymous
    • 5 years ago
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    http://www.wolframalpha.com/input/?i=y%27%27%27%27-3y%27%27%27%2B4y%27%27+%3D+2%28x-1%29e^x

  2. anonymous
    • 5 years ago
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    lol ya ok... could've done that... need the work

  3. anonymous
    • 5 years ago
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    what is 4"

  4. anonymous
    • 5 years ago
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    This is a massive question to do online. I'll do as much as I can...

  5. anonymous
    • 5 years ago
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    Make the substitution \[v=y'\]Then your equation becomes\[v''-3v'+4v=e^x(2x-2)\]

  6. anonymous
    • 5 years ago
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    You need to solve the homogeneous equation first, then find the particular solution.

  7. anonymous
    • 5 years ago
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    \[v''-3v'+4v=0\]has characteristic equation (assuming a solution \[v=e^{{\lambda}x}\])\[{\lambda}^2-3{\lambda}+4=0\rightarrow {\lambda}=\frac{3}{2}{\pm}\frac{\sqrt{7}}{2}i\]which yields a homogeneous solution,\[v=c_1e^{3x/2}\cos(\frac{\sqrt{7}x}{2})+c_2e^{3x/2}\sin(\frac{\sqrt{7}x}{2})\]

  8. anonymous
    • 5 years ago
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    The particular solution can be found by attempting a trial solution based on the RHS of the DE, so attempt\[v=e^x(ax+b)\]When you substitute this into the DE, you end up with the following\[2ax+(2b-a)=2x-2\](the exponentials cancel). This is true only for a=1 and b=-1/2, so the particular solution is\[v=e^x(x-\frac{1}{2})\]

  9. anonymous
    • 5 years ago
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    Your total solution is therefore,

  10. anonymous
    • 5 years ago
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    \[v=c_1e^{3x/2}\cos(\frac{\sqrt{7}}{2}x)+c_2e^{3x/2}\sin(\frac{\sqrt{7}}{2}x)+e^x(x-\frac{1}{2})\]

  11. anonymous
    • 5 years ago
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    Now, since \[v=\frac{dy}{dx}\] you have to integrate v to find y...this I shall leave to you...if you need further assistance, let me know. Good luck.

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