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Homework Help: First Order Linear Differential Equations

  1. Jan 18, 2004 #1

    It has been over a year since I last did calculus. And I am having trouble with my current calculus course. First here is the question:

    Solve the given first-order linear equation and verify that your solution indeed satisfies the equation.

    y' - 2xy = 2xe^x^2

    Now I THINK I have the answer:

    y = e^x^2 ( x^2 + c)

    But how do I verify? I would think I simply would take the above equation and it's derivative and sub. into the equation

    y' - 2xy = 2xe^x^2

    If that is the case, my problem is this: When I take the derivative of

    y = e^x^2 ( x^2 + c)

    I have

    y' = 2x * e^x^2 + 2x^3 * e^x^2 + 2x * C * e^x^2

    How can I verify my answer when there is an unknown constant in my derivative? What am I missing?

    Any help is appreciated - detailed if possible. Thankyou.
  2. jcsd
  3. Jan 18, 2004 #2


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    Keep doing what you're doing and the constant will cancel out:

    [tex] y = e^{x^2}(x^2+{\rm C}) [/tex]
    [tex] y' = 2x{\rm C}e^{x^2} + 2x^3e^{x^2} + 2xe^{x^2} [/tex]
    [tex] 2xy = 2x{\rm C}e^{x^2} + 2x^3e^{x^2} [/tex]

    Substitute in to find that:

    [tex] y'-2xy = 2xe^{x^2} [/tex]
  4. Jan 18, 2004 #3


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    Ok, as I understand it we have the following equation that we must solve.

    [tex] \frac {dy}{dx}-2xy=2xe^{x^2} [/tex]

    First we must compute the integrating factor which is.

    [tex] F(x)=e^{\int{f(x)dx}}[/tex]

    and multiply the original equation by this to get

    [tex] e^{x^2}\frac{dy}{dx}-2xye^{x^2}=2xe^{2x^2}[/tex]

    And then the final equation to solve just becomes


    and y is then

    [tex] y=-\frac{e^{2x^2}}{2}+C[/tex]
    Last edited: Jan 18, 2004
  5. Jan 18, 2004 #4
    Thanks to the both of you.

    I think kurdt that you made an error.

    f(x) = - 2x


    f(x) = 2x

    Thanks anyway though.

  6. Jan 19, 2004 #5


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    Oops I apologise. Well it just goes to prove I am only human :smile:
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