Diff EQ

F

Frogger888

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Feb 2, 2006
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We just started going over solving systems of linear equations by elimination and I am having a tough time getting started
This is the problem
dx/dt=4x+7
dy/dt=x-2

What I have done so far is convert to

Dx-4x-7y
Dy-x+2y

next

Dx-3x-7y=0
D(Dy-x+2y)=0=DDy-Dx+2Dy now my Dx cancels and I get

DDy-3x+2Dy-7y=0 which yields

DDy+2Dy-7y=3x next

Somewhere before this sentence I feel I have made a mistake; Can someone check it out and let me know if I did something wrong' Like I said I am just learning this new method Thanks

m^2+2m-7=0 gives me a imaginery number and I do the normal steps to give me a value then I annhilate the right side (3x) using DD and solve for y partial
 
Do you mean:

\(\displaystyle \L\\\frac{dx}{dt}=4x+7y\)
\(\displaystyle \L\\\frac{dy}{dt}=x-2y\)?.

I believe you left off your 'y's' in the beginning.

Anyway,

From \(\displaystyle \L\\D_{x}=4x+7y\) and \(\displaystyle \L\\D_{y}=x-2y\) we get

\(\displaystyle \L\\y=\frac{1}{7}D_{x}-\frac{4}{7}x\) and

\(\displaystyle \L\\D_{y}=\frac{1}{7}D^{2}_{x}-\frac{4}{7}D_{x}\)

\(\displaystyle \L\\x=D_{y}+2y\)

\(\displaystyle \L\\D_{x}=D^{2}y+2Dy\)

Now, can you find the auxiliary equation?.
 
I understand how you got some of that information but exactly what am I supposed to do with it. I am really lost on this whole concept of problems

I am getting

DDx/7-D4x/7+2y=0

from making it to where I can subract the information from the two equations.
But what now I have a y term in with these x terms and we have done no examples showing us how to solve something of this nature
 
Hey I finally understand this stuff I just had to take a break and think about it
Thanks for the help
Frogger
 
Great!. Sometimes it helps to take a break. It's nice to see a light bulb come on:)
 
Okay I thought I got it but I am wrong

I am getting for my values

DDy+2yD-7y-4Dy-8y=0

that will give me my y(t) when I solve it out I am getting

y(t)=ce^(6t)+ce^(-2t)

Any advice
 
I believe you should get \(\displaystyle (D^{2}-2D-15)x\)., if I recollect right.
 
That is what I get
(DD-2D-15)y;
Then I solve with a quadratic and get D=6 m=-2

x=4+ or - square root of [4+64] all over 2

x=2 + or - 4 x=6, -2

y(t)=c1e^(6t) +c2e^(-2t)

does this look correct. Next I must solve for x(t) correct?
 
You're getting the wrong answer in your quadratic.

\(\displaystyle m^{2}-2m-15=0\)

\(\displaystyle m^{2}-2m=15\)

\(\displaystyle (m^{2}-2m+1)=15+1\)

\(\displaystyle (m-1)^{2}=16\)

\(\displaystyle m-1=\pm{4}\)

m=-3,5
 
I thought that did not seem right thank you so much for the help. I really do appreciate it
Frogger
 
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