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 Differential Equations Ordinary and Partial Differential Equations Math Forum

 November 21st, 2016, 08:54 AM #1 Newbie   Joined: Nov 2016 From: nottingham, uk Posts: 3 Thanks: 0 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 Last edited by skipjack; November 21st, 2016 at 09:40 PM. November 21st, 2016, 09:46 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,685 Thanks: 2665 Math Focus: Mainly analysis and algebra 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. Thanks from topsquark Last edited by v8archie; November 21st, 2016 at 10:06 AM. November 21st, 2016, 07:18 PM   #3
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Quote:
 Originally Posted by chris9598 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*} November 21st, 2016, 10:14 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,975 Thanks: 2226 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? November 22nd, 2016, 12:07 AM #5 Newbie   Joined: Nov 2016 From: nottingham, uk Posts: 3 Thanks: 0 hi, yeah thats the correct one. i got this from a problem sheet in college November 22nd, 2016, 03:05 AM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,685 Thanks: 2665 Math Focus: Mainly analysis and algebra $$(6e)^{4t-9}=e^{-9} e^{(1+\ln 6)4t}$$ There's nothing too unusual here, we just have awkward values for constants. November 22nd, 2016, 06:07 AM   #7
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Did you mean $(6e)^{-9} e^{(1+\ln 6)4t}$, v8archie?

Quote:
 Originally Posted by chris9598 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? November 22nd, 2016, 07:53 AM #8 Newbie   Joined: Nov 2016 From: nottingham, uk Posts: 3 Thanks: 0 The right-hand side should be 6e^(4t)-9 Last edited by skipjack; November 22nd, 2016 at 08:28 AM. November 22nd, 2016, 07:53 AM #9 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,685 Thanks: 2665 Math Focus: Mainly analysis and algebra 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. November 22nd, 2016, 07:54 AM   #10
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Quote:
 Originally Posted by chris9598 the right hand side should be 6e^(4t)-9
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