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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) = ?
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) = ?

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Make 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
 5 years ago
Best ResponseYou've already chosen the best response.0With 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,\[(\lambda 3)(\lambda2)=0 \rightarrow \lambda = 2, 3\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Your homogeneous solution in v is then\[v=c_1e^{2x}+c_2e^{3 x}\]But v = y', so\[y'=c_1e^{2x}+c_2e^{3 x} \rightarrow y = c_1e^{2x}+c_2 e^{3x}+c_3\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The constants on the far righthand side have just absorbed the constants from integrating the exponentials.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can then solve with your boundary conditions.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[y(0)=c_1+c_2+c_3=0\]\[y'=2c_1e^{2x}+3c_2e^{3x} \rightarrow y'(0)=2c_1+3c_2=3\] (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
 5 years ago
Best ResponseYou've already chosen the best response.0Working through the simultaneous equations quickly, I get\[c_1=\frac{9}{2},c_2=4, c_3=\frac{1}{2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks, 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
 5 years ago
Best ResponseYou've already chosen the best response.0Okay...well, that's how you do it. Check the constants. Feel free to give me a fan point :p
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