solving linear ode

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  • 7/25/2019 Solving Linear ODE

    1/2

    2/2/2016 Solving Linear ODE Using Laplace Transforms

    http://math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/ode/laplace/solve/solve.html 1/2

    Solving Linear ODE Using Laplace Transforms

    How canwe use Laplace transforms to solve ode? The procedure is bestillustrated with an example. Consider the ode

    This is a linear homogeneous ode and can be solved using standard methods.

    Let Y(s)=L[y(t)](s). Instead of solving directly for y(t), we derive anew equation for Y(s). Once we find Y(s), we inverse transformto determine y(t).

    The first step is to take the Laplace transform of both sides of theoriginal differential equation. We have

    Obviously, the Laplace transform of the function 0 is 0. If we look atthe left-hand side, we have

    Now use the formulas for the L[y'']and L[y']:

    Here we have used the fact that y(0)=2. And,

    Hence, we have

    The Laplace-transformed differential equation is

    This is a linear algebraic equation for Y(s)! We have converted adifferential equation into a algebraic equation! Solving for Y(s), we have

    We can simplify this expression using the method of partial fractions:

    Recall the inverse transforms:

    Using linearityof the inverse transform, we have

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    2/2/2016 Solving Linear ODE Using Laplace Transforms

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    Another Example

    Consider the ode:

    This problem has an inhomogeneous term. In the direct approachone solves for the homogeneous solution and the particular solutionseparately. For this problem the particular solution can be determinedusing variation of parameters or the method of undeterminedcoefficients. Using the Laplace transform technique we can solve forthe homogeneous and particular solutions at the same time.

    Let Y(s) be the Laplace transform of y(t). Taking the Laplace transformof the differential equation we have:

    The Laplace transform of the LHS L[y''+4y'+5y] is

    The Laplace transform of the RHS is

    Equating the LHS and RHS and using the fact that y(0)=1 y'(0)=2, we

    obtain

    Solving for Y(s), we obtain:

    Using the method of partial fractionsit can be shown that

    Using the fact that the inverseof 1/(s-1) is e^t and that the inverseof1/[(s+2)^2+1] is exp(2t)sin(t), it follows that

    [ODE Home] [1st-Order Home] [2nd-Order Home] [Laplace Transform Home] [Notation]

    [References]

    Copyright 1996 Department of Mathematics, Oregon State University

    If you have questions or comments, don't hestitate to contact us.

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