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Math Help - Systems of Equations

  1. #1
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    Systems of Equations

    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
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  2. #2
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    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.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    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
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  4. #4
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    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...
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  5. #5
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    Quote Originally Posted by Nguyen View Post

    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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