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anonymous

  • 5 years ago

Find y as a function of x if: y''' - 5y'' + 6y' = 0 y(0) = 2, y'(0) = 3, y''(0) = 0 y(x) = ?

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  1. anonymous
    • 5 years ago
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    Make a substitution, v=y'. Then form, v'' - 5v' + 6v = 0 which is something I'm thinking you know how to solve. Once you've found the two homogeneous solutions, you have to undo your substitution.

  2. anonymous
    • 5 years ago
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    With your 'v' equation, assume it has solution of the form\[v=e^{\lambda x}\]Then\[v''-5v+6=(e^{\lambda x})'-5e^{\lambda x}+6=e^{\lambda x}(\lambda ^2 -5 \lambda +6)=0\]Since \[e^{\lambda x} \ne 0\]for any x, it must be that the quadratic in lambda is equal to zero. Factoring gives,\[(\lambda -3)(\lambda-2)=0 \rightarrow \lambda = 2, 3\]

  3. anonymous
    • 5 years ago
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    Your homogeneous solution in v is then\[v=c_1e^{2x}+c_2e^{3 x}\]But v = y', so\[y'=c_1e^{2x}+c_2e^{3 x} \rightarrow y = c_1e^{2x}+c_2 e^{3x}+c_3\]

  4. anonymous
    • 5 years ago
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    The constants on the far right-hand side have just absorbed the constants from integrating the exponentials.

  5. anonymous
    • 5 years ago
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    You can then solve with your boundary conditions.

  6. anonymous
    • 5 years ago
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    \[y(0)=c_1+c_2+c_3=0\]\[y'=2c_1e^{2x}+3c_2e^{3x} \rightarrow y'(0)=2c_1+3c_2=3\] (here we do care about the constants and don't want parts differentiated to be absorbed, since we care about the relationship between this equation and the first one. The same will be the case for y'').\[y''=4c_1e^{2x}+9c_2e^{3x} \rightarrow y''(0)=4c_1+9c_2=0\]

  7. anonymous
    • 5 years ago
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    Working through the simultaneous equations quickly, I get\[c_1=-\frac{9}{2},c_2=4, c_3=\frac{1}{2}\]

  8. anonymous
    • 5 years ago
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    Hello..?

  9. anonymous
    • 5 years ago
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    Thanks, I understand how to solve it but we did not cover this in our differential class yet. I appreciate all the help you have given.

  10. anonymous
    • 5 years ago
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    Okay...well, that's how you do it. Check the constants. Feel free to give me a fan point :p

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