# Systems of Equations

• May 13th 2011, 07:38 PM
Nguyen
Systems of Equations
Hey, for this System I want to find the general solution but $\displaystyle y_1 '$ and $\displaystyle y_2 '$ both have constants which I can't remember how to deal with.

Find the general solution to:

$\displaystyle y_1 ' =-y_1 +2$

$\displaystyle y_2 ' = -y_2 -1$

And I have a non-linear system that needs to be linearised but my working is getting me nowhere.

Linearise the system around the critical point (0,0):

$\displaystyle \frac{dx}{dt}=1-e^{x+y}$

$\displaystyle \frac{dy}{dt}=2x+sin(y)$

Any help would be greatly appreciated.

Thanks
• May 13th 2011, 07:46 PM
Prove It
Rewrite each equation as $\displaystyle \displaystyle y_1' + y_1 = 2$ and $\displaystyle \displaystyle y_2' + y_2 = -1$.

Each of those is first-order linear, so solve each using the integrating factor method.
• May 13th 2011, 07:58 PM
Nguyen
Quote:

Originally Posted by Prove It
Rewrite each equation as $\displaystyle \displaystyle y_1' + y_1 = 2$ and $\displaystyle \displaystyle y_2' + y_2 = -1$.

Each of those is first-order linear, so solve each using the integrating factor method.

Thanks. So I get $\displaystyle y_1 = 2x + \frac{c}{e^x}$, $\displaystyle y_2 = -x + \frac{c}{e^x}$. So what do I do now? Sorry for the noob question, haven't solved systems like this before.

Any thoughts on the second question?

Thanks
• May 13th 2011, 08:18 PM
Prove It
I get $\displaystyle \displaystyle y_1 = 2 + \frac{c_1}{e^x}$ and $\displaystyle \displaystyle y_2 = -1 + \frac{c_2}{e^x}$ - remember that when you multiply by the integrating factor, you have to multiply it to BOTH sides of the equation. Also, the arbitrary constants might not be the same... And surely that's the solution to the system of DEs since you don't have any boundary conditions...
• May 13th 2011, 08:34 PM
TheEmptySet
Quote:

Originally Posted by Nguyen

Any thoughts on the second question?

Thanks

Given a system of the form

$\displaystyle \frac{dx}{dt}=P(x,y)$

$\displaystyle \frac{dy}{dt}=Q(x,y)$

If we expand P and Q in a Taylor series at (0,0) to first order terms we get

$\displaystyle \frac{dx}{dt}=P(x,y) \approx P(0,0)+\frac{\partial P}{\partial x}\bigg|_{(0,0)}(x-0)+ \frac{\partial P}{\partial y}\bigg|_{(0,0)}(y-0)$

$\displaystyle \frac{dy}{dt}=Q(x,y) \approx Q(0,0)+\frac{\partial Q}{\partial x}\bigg|_{(0,0)}(x-0)+ \frac{\partial Q}{\partial y}\bigg|_{(0,0)}(y-0)$

Since (0,0) is a critical point we have that

$\displaystyle P(0,0)=Q(0,0)=0$

Plugging the the appropriate partials gives

$\displaystyle \frac{dx}{dt}=-x-y$

$\displaystyle \frac{dy}{dt}= 2x +y$

Also on the first problem are you sure it was typed correctly and is not

$\displaystyle y_1'=-y_2+2 \quad y_2 = -y_1-1$

otherwise it is not much of a system.
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