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Thread: the form of the solution of a linear ODE

  1. #1
    No one in Particular VonNemo19's Avatar
    Apr 2009
    Detroit, MI

    the form of the solution of a linear ODE

    Hi everyone.

    I'm having a little trouble understanding something, so maybe (hopefully) you guys can help.

    It is a well known fact in differential equations that the solution to a first order linear differential equation is the sum of the solution to the associated homogeneous equation and the solution to the particular equation, that is

    $\displaystyle y=y_c+y_p$, where $\displaystyle y_c$ is the solution of

    $\displaystyle \frac{dy}{dx}+P(x)y=0$

    and $\displaystyle y_p$ of

    $\displaystyle \frac{dy}{dx}+P(x)y=f(x) $

    Now, I understand that this is a truism because

    $\displaystyle \frac{d}{dx}[y_c+y_p]+P(x)[y_c+y_p]=\underbrace{\frac{dy_c}{dx}+P(x)y_c}_0+\underbrac e{\frac{dy_p}{dx}+P(x)y_p}_{f(x)}=f(x)$,

    but what I'm having trouble with is that of course if I add zero to anything it won't change the element at all because of the fact that we are working on the real numbers here. So, what is the purpose here? I mean, I could easily say that...

    The number 5 has the property that it is the sum

    $\displaystyle \frac{dy_c}{dx}+P(x)y_c+5=5$.

    But to what end?

    By the way. I'm really sorry if this question makes no sense to you. If it doesn't, just tell me and I'll try to figure out what my malfunction is on my own.

    Thank you all for reading.
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  2. #2
    A Plied Mathematician
    Jun 2010
    CT, USA
    Here's the purpose: it's all about IVP's. If you have initial conditions, your particular solution isn't necessarily going to satisfy them. But, with the homogeneous solution, which contains arbitrary constants of integration, you can tailor your solution to match the initial condition. That's why you want the homogeneous part in there.
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