November 21st, 2016, 08:54 AM  #1 
Newbie Joined: Nov 2016 From: nottingham, uk Posts: 3 Thanks: 0  ODEs
dx/dt+xy=1 dy/dt+27x9y=(6e)^4t9 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: 7,671 Thanks: 2651 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. Last edited by v8archie; November 21st, 2016 at 10:06 AM. 
November 21st, 2016, 07:18 PM  #3  
Member Joined: Oct 2016 From: Melbourne Posts: 77 Thanks: 35  Quote:
$\displaystyle \begin{align*} \frac{\mathrm{d}x}{\mathrm{d}t} + x  y &= 1 \\ \frac{\mathrm{d}y}{\mathrm{d}t} + 27\,x  9\,y &= \left( 6\,\mathrm{e} \right) ^{4\,t}  9 \end{align*}$  
November 21st, 2016, 10:14 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,831 Thanks: 2160 
Was your second equation intended to be dy/dt + 27x  9y = (6e)^(4t9), 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: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra 
$$(6e)^{4t9}=e^{9} e^{(1+\ln 6)4t}$$ There's nothing too unusual here, we just have awkward values for constants. 
November 22nd, 2016, 06:07 AM  #7 
Global Moderator Joined: Dec 2006 Posts: 20,831 Thanks: 2160  
November 22nd, 2016, 07:53 AM  #8 
Newbie Joined: Nov 2016 From: nottingham, uk Posts: 3 Thanks: 0 
The righthand 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: 7,671 Thanks: 2651 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 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra  

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