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Thread: Linear , homogenous, separation and change of variables

  1. #31
    MHF Contributor MarkFL's Avatar
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    Re: Linear , homogenous, separation and change of variables

    Can you show whether it is exact or not?
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  2. #32
    Junior Member romeobernard's Avatar
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    Re: Linear , homogenous, separation and change of variables

    My = x + 1
    Nx = 6x - 1

    Is this right?

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  3. #33
    Junior Member romeobernard's Avatar
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    Re: Linear , homogenous, separation and change of variables

    I mean not exact right..?

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    Junior Member romeobernard's Avatar
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    Re: Linear , homogenous, separation and change of variables

    (x+y)dx + (3x^2 - 1)dy = 0

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  5. #35
    MHF Contributor MarkFL's Avatar
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    Re: Linear , homogenous, separation and change of variables

    The equation is not exact, but check your partial differentiation...
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  6. #36
    Junior Member romeobernard's Avatar
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    Re: Linear , homogenous, separation and change of variables

    Nx= 6x?

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  7. #37
    MHF Contributor MarkFL's Avatar
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    Re: Linear , homogenous, separation and change of variables

    $\displaystyle M_y(x,y)=1$

    $\displaystyle N_y(x,y)=6x$

    Since these are not equal, the equation is not exact. So, what should we do to seek an integrating factor?

    Just to be clear, the equation is actually:

    $\displaystyle (x+y)dx+(3x^2-1)dy=0$

    Correct?
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  8. #38
    Junior Member romeobernard's Avatar
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    Re: Linear , homogenous, separation and change of variables

    Yes. Its

    (x+y)dx + (3x^2 - 1)dy = 0

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  9. #39
    MHF Contributor MarkFL's Avatar
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    Re: Linear , homogenous, separation and change of variables

    Quote Originally Posted by romeobernard View Post
    Yes. Its

    (x+y)dx + (3x^2 - 1)dy = 0
    Okay, good...that makes finding an integrating factor possible. Can you find such a factor?
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  10. #40
    Junior Member romeobernard's Avatar
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    Re: Linear , homogenous, separation and change of variables

    I think there is no way we can find integrating factor base on the standard linear form.

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  11. #41
    MHF Contributor MarkFL's Avatar
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    Re: Linear , homogenous, separation and change of variables

    Quote Originally Posted by romeobernard View Post
    I think there is no way we can find integrating factor base on the standard linear form.
    According to the theory I was taught as a student, we should try:

    $\displaystyle \mu(x)=\exp\left(\int \frac{1-6x}{3x^2-1}\,dx\right)=\,?$
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  12. #42
    Junior Member romeobernard's Avatar
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    Re: Linear , homogenous, separation and change of variables

    Quote Originally Posted by MarkFL View Post
    According to the theory I was taught as a student, we should try:

    $\displaystyle \mu(x)=\exp\left(\int \frac{1-6x}{3x^2-1}\,dx\right)=\,?$
    Ah..yes i see it its the IF for non exact equation.

    But the next step is using a partial fraction differentiation right?

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  13. #43
    MHF Contributor MarkFL's Avatar
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    Re: Linear , homogenous, separation and change of variables

    Yes, partial fraction decomposition on the integrand would be a good idea, so you can get the integral in terms of natural logs.
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  14. #44
    Junior Member romeobernard's Avatar
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    Re: Linear , homogenous, separation and change of variables

    Mark how about this one.I dont know where to start or what DE i will use.

    (x+2y-1)dx - (2x+y-5)dy = 0


    Thanks.

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  15. #45
    MHF Contributor MarkFL's Avatar
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    Re: Linear , homogenous, separation and change of variables

    What I would do first is solve the system:

    $\displaystyle x+2y-1=0$

    $\displaystyle 2x+y-5=0$

    From which we obtain:

    $\displaystyle (x,y)=(3,-1)$

    And so let's let:

    $\displaystyle x=u+3\implies dx=du$

    $\displaystyle y=v-1\implies dy=dv$

    And the ODE becomes:

    $\displaystyle ((u+3)+2(v-1)-1)dx-(2(u+3)+(v-1)-5)dy=0$

    $\displaystyle (u+2v)du-(2u+v)dv=0$

    $\displaystyle \frac{dv}{du}=\frac{u+2v}{2u+v}=\frac{1+2\dfrac{v} {u}}{2+\dfrac{v}{u}}$

    Now you have a first order homogeneous equation...can you proceed?
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