anonymous
  • anonymous
Find y as a function of x if: y''' - 5y'' + 6y' = 0 y(0) = 2, y'(0) = 3, y''(0) = 0 y(x) = ?
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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.
anonymous
  • anonymous
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,\[(\lambda -3)(\lambda-2)=0 \rightarrow \lambda = 2, 3\]
anonymous
  • anonymous
Your homogeneous solution in v is then\[v=c_1e^{2x}+c_2e^{3 x}\]But v = y', so\[y'=c_1e^{2x}+c_2e^{3 x} \rightarrow y = c_1e^{2x}+c_2 e^{3x}+c_3\]

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anonymous
  • anonymous
The constants on the far right-hand side have just absorbed the constants from integrating the exponentials.
anonymous
  • anonymous
You can then solve with your boundary conditions.
anonymous
  • anonymous
\[y(0)=c_1+c_2+c_3=0\]\[y'=2c_1e^{2x}+3c_2e^{3x} \rightarrow y'(0)=2c_1+3c_2=3\] (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'').\[y''=4c_1e^{2x}+9c_2e^{3x} \rightarrow y''(0)=4c_1+9c_2=0\]
anonymous
  • anonymous
Working through the simultaneous equations quickly, I get\[c_1=-\frac{9}{2},c_2=4, c_3=\frac{1}{2}\]
anonymous
  • anonymous
Hello..?
anonymous
  • anonymous
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.
anonymous
  • anonymous
Okay...well, that's how you do it. Check the constants. Feel free to give me a fan point :p

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