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Math Help - Seperation of variables/Linear Differential

  1. #1
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    Seperation of variables/Linear Differential

    I am right with my assumptions regarding these equations?

    x3(dy/dx) + 2x^2y = 5 -----> use linear differential to solve equation?

    dy/dt = 1 +y^2/y(1+t^2) ------> use the separating the variables method? also in this one can i get rid of the brackets on the bottom to leave me with y + t^2?

    cheers
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  2. #2
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    For the separating variables equation I have:

    g(t) = 1 + y^2
    h(t) = y + t^2

    So

    1/y+t^2 (dy/dt) = 1/y^2


    (Integration) 1/y+t^2 (dy/dt) dt = (Integration) 1 +y^2 dt

    (Integration) 1/y+t^2 dt = (Integration) 1 +y^2 dt

    Arctan t = ????

    What is the integral for 1 + y^2?

    and am I along the right lines?
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  3. #3
    Member Altair's Avatar
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    Quote Originally Posted by poundedintodust View Post
    x3(dy/dx) + 2x^2y = 5 -----> use linear differential to solve equation?
    This is a homogeneous equation. Simply do it by substitution. Take y=Vx. Then separate the variables.
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  4. #4
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    sorry to be ignorant but could you give a bit more of an explanation? just this is all new to me!!!!
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  5. #5
    Member Altair's Avatar
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    I am sorry. You can't do it by that method. I didn't see that 5. It isn't a homogeneous equation.
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  6. #6
    Member Altair's Avatar
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    I have yet to do non-homogeneous equations. I am a noobest noob in ODEs.
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  7. #7
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    Quote Originally Posted by poundedintodust View Post
    I am right with my assumptions regarding these equations?

    x3(dy/dx) + 2x^2y = 5 -----> use linear differential to solve equation?
    Written backwards as y' + \frac{{2y}}<br />
{x} = \frac{5}<br />
{{x^3 }}, and this is a linear ODE, you can tackle it by using the integrating factor method.

    As for your 2nd. question, the equation is separable. Take it to the form f(y)\,dy=f(t)\,dt.
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