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 chris9598 November 21st, 2016 08:54 AM

ODEs

dx/dt+x-y=-1
dy/dt+27x-9y=(6e)^4t-9

Find x(t) and y(t) subject to the initial conditions x(0)=3, y(0)=18

 v8archie November 21st, 2016 09:46 AM

Your complementary solution to $$\mathbf{x}'=\mathbf{Ax} \quad \text{where} \quad \mathbf{A}=\begin{pmatrix}-1 & 1 \\ -27 & 9 \end{pmatrix}$$
will be augmented by a particular solution which will depend on the eigenvalues of $\mathbf{A}$, but may well be $$\mathbf{x}_p=\mathbf{c}_1 e^{4t} + \mathbf{c}_2$$
where the vectors $\mathbf{c}_1$ and $\mathbf{c}_2$ can be determined by the method of undetermined coefficients.

 Prove It November 21st, 2016 07:18 PM

Quote:
 Originally Posted by chris9598 (Post 554143) dx/dt+x-y=-1 dy/dt+27x-9y=〖6e〗^4t-9 Find x(t) and y(t) subject to the initial conditions x(0)=3, y(0)=18
The way you have written the equations is ambiguous. Is this what you meant?

\displaystyle \begin{align*} \frac{\mathrm{d}x}{\mathrm{d}t} + x - y &= -1 \\ \frac{\mathrm{d}y}{\mathrm{d}t} + 27\,x - 9\,y &= \left( 6\,\mathrm{e} \right) ^{4\,t} - 9 \end{align*}

 skipjack November 21st, 2016 10:14 PM

Was your second equation intended to be dy/dt + 27x - 9y = (6e)^(4t-9), chris9598? Also, where did you get this problem from?

What were the characters enclosing 6e that I've replaced with parentheses?

 chris9598 November 22nd, 2016 12:07 AM

hi, yeah thats the correct one. i got this from a problem sheet in college

 v8archie November 22nd, 2016 03:05 AM

$$(6e)^{4t-9}=e^{-9} e^{(1+\ln 6)4t}$$
There's nothing too unusual here, we just have awkward values for constants.

 skipjack November 22nd, 2016 06:07 AM

Did you mean $(6e)^{-9} e^{(1+\ln 6)4t}$, v8archie?

Quote:
 Originally Posted by chris9598 (Post 554200) yes, that's the correct one. I got this from a problem sheet in college.
Just to confirm, the right-hand side should be (6e)^(4t-9), not (6e)^(4t) - 9, and not 6e^(4t - 9) either?

 chris9598 November 22nd, 2016 07:53 AM

The right-hand side should be 6e^(4t)-9

 v8archie November 22nd, 2016 07:53 AM

Oh, yeah. I missed a bit. It's still $ae^{kt}$, just with a very messy $k$. That $a$ will get swallowed up in constants that we find using the method of undetermined coefficients.

 v8archie November 22nd, 2016 07:54 AM

Quote:
 Originally Posted by chris9598 (Post 554233) the right hand side should be 6e^(4t)-9