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Simple differential equation substitution

  1. Jan 17, 2007 #1

    I am looking through some solved exercises. One of them is the following:

    Solve the equation: x^2 y'' + (x^2 - 3x)y' + (3-x)y = x^4
    knowing that y=x is a solution of the homogeneous equation.

    The professor then solves it by doing the following substitution: y=xz.
    Then he calculates y' and y'', substitutes, etc.

    What I do not understand is, how do you know, from the fact that y=x is a solution to the homogeneous equation, that you have to do that substitution y=xz?
  2. jcsd
  3. Jan 17, 2007 #2


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    What is z?
  4. Jan 17, 2007 #3


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    You don't "know" that- you don't have to do that substitution- there are many ways to solve a differential equation. However, it is true that what ever function the correct solution y is, there exist a function z such that y= xz: z= y/x, of course.

    The point is this: if y= f(x)z (for ANY function f(x) that satisfies the equation), then y'= f'(x)z+ f(x)z', y"= f"(x)z+ 2f'(x)z'+ f(x)z". I've used the product rule twice here. Notice that, in the terms that involve only z, not z' or z", I have differentiated only the f(x) term. It is exactly as if z were a constant. Since f(x) satisfies this linear homogenous equation so does any constant times f(x): the terms of involving only z wil all cancel out leaving a differential equation for z' and higher derivatives. Now let u(x)= z' and you have a differential equation for u of lower order.

    To repeat: you don't have to do that substitution- but it can be done and has a good chance of simplifying the problem so it is worth trying!
  5. Jan 17, 2007 #4
    I see.
    So if I were told that y=x^2 was a solution to the homogeneous eq, then I should try y=x^2z?
  6. Jan 17, 2007 #5


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    Yes, exactly. That's known as "reduction of order": x^2 is a solution to an nth order linear homogeneous diff eq, then y= x^2z will be a solution for some z satisfying an n-1 order equation. It's very similar to reducing the degree of a polynomial by dividing by x-a if you already know that a is a solution.
  7. Jan 31, 2007 #6
    Just solve it using a power series. It's far easier to do that then to try and find substitutions every single time.
  8. Jan 31, 2007 #7


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    What "substitutions" are you talking about? The question here was about using the fact that you know one solution to a differential equation to reduce the order.
  9. Jan 31, 2007 #8
    I'm saying that if you know y = xz, then for every variable in the equation you are trying to substitute to reduce to a single variable. While that is helpful, it doesn't always yield a solution to the equation. Therefore, in instances when you don't know y = xz, it would be wise to solve using a power series.

    Perhaps its just the fact that I love power series that it would come more natural for me to do so in that manner.
  10. Feb 1, 2007 #9


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    The method of variation of constants. [itex] y_{p}(x)=C(x) y_{o}(x)=C(x) x=z(x)x=zx [/itex].
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