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Solving a linear homogenous 2nd order DE with a constant term

  1. Aug 15, 2009 #1
    I know how to go about solving differential equations of the form y''+q(x)y'+t(x)y = 0 through the methods of finding the characteristic polynomial of the differential equation and solving for the roots, etc. But what I am not clear on is how I would go about solving an equation like this where one of the terms is a constant, such as y''+q(x)y' + C = 0 or y'' +t(x)y + C = 0. I'm thinking of a situation with an F=ma equation where there is a term that is dependent on the position or velocity of an object, but there is also something like a gravitational force, which is constant.
     
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  3. Aug 15, 2009 #2

    Pengwuino

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    Gold Member

    Do you mean that your q(x) and t(x) are constants? The method you described doesn't work if they are not constants. If you simply have a constant or some term that doesn't have y,y', or y'', your equation is simply inhomogeneous. You tack on a particular solution that depends on what the inhomogeneous part is. If you have a inhomogeneous part, the full solution is simply [tex]y(x) = y_c + y_p [/tex] where [tex] y_c[/tex] is the solution to the homogeneous equation and [tex] y_p[/tex] is the solution to the inhomogeneous equation. For a constant inhomogeneous part, you try [tex]y_p = C[/tex] where C is a constant. You plug this into the DE and solve for it. If the solution for your homogeneous equation is a constant, you can't use that for a particular solution. You can try [tex]y_p = Ax + B[/tex] and solve for A and B to form your solution.
     
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