Systems of Equations

**Nguyen** · May 13th 2011, 07:38 PM

Hey, for this System I want to find the general solution but $y_1 '$ and $y_2 '$ both have constants which I can't remember how to deal with.

Find the general solution to:

$y_1 ' =-y_1 +2$

$y_2 ' = -y_2 -1$

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):

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

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

Any help would be greatly appreciated.

Thanks

**Prove It** · May 13th 2011, 07:46 PM

Rewrite each equation as $\displaystyle y_1' + y_1 = 2$ and $\displaystyle y_2' + y_2 = -1$ .

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

**Nguyen** · May 13th 2011, 07:58 PM

Originally Posted by Prove It

Rewrite each equation as $\displaystyle y_1' + y_1 = 2$ and $\displaystyle y_2' + y_2 = -1$ .

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

Thanks. So I get $y_1 = 2x + \frac{c}{e^x}$ , $y_2 = -x + \frac{c}{e^x}$ . 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

**Prove It** · May 13th 2011, 08:18 PM

I get $\displaystyle y_1 = 2 + \frac{c_1}{e^x}$ and $\displaystyle y_2 = -1 + \frac{c_2}{e^x}$ - 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...

**TheEmptySet** · May 13th 2011, 08:34 PM

Originally Posted by Nguyen

Any thoughts on the second question?

Thanks

Given a system of the form

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

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

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

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

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

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

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

Plugging the the appropriate partials gives

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

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

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

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

otherwise it is not much of a system.

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