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Math Help - Verifying solutions for differential equations

  1. #1
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    Verifying solutions for differential equations

    Verify this is a solution of this differential equation:

    S=integrand......integrate from 0 to x
    dy/dx+2xy = 1 ; y = e^(-x^2)Se^(t^2) dt +c[1]e^(-x^2)

    [Difficulty]
    I'm not sure how to simplify y. It looks like the fundamental theorem
    of calculus. So e^(-x^2)S(from 0 to x)e^(t^2)dt should just be 1,
    right? I let y=1+c[1]e^(-x^2) and used implicit differentiation, but
    I end up with dy/dx+2xy=2x. If 2x=1 then I'd have it. I also
    substituted y and dy/dx into the equation, but I get the same thing:
    2x=1. I have to be simplifying y wrong. Can you help me? Thanks!

    [Thoughts]
    y=1+ce^(-x^2)
    (y-1)/e^(-x^2)=c
    e^(x^2)y-e^(x^2)=c
    e^(x^2)y'+2xye^(x^2)-2xe^(x^2)=0
    e^(x^2)y'=2xe^(x^2)-2xye^(x^2)
    y'=2x-2xy
    y'+2xy=2x

    dy/dx+2xy=1; y=1+ce^(-x^2); dy/dx=-2xce^(-x^2)
    -2xce^(-x^2)+2x(1+ce^(-x^2))=1
    -2xce^(-x^2)+2x+2xce^(-x^2)=1
    2x=1
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  2. #2
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    Quote Originally Posted by duaneg37 View Post
    Verify this is a solution of this differential equation:

    S=integrand......integrate from 0 to x
    dy/dx+2xy = 1 ; y = e^(-x^2)Se^(t^2) dt +c[1]e^(-x^2)

    [Difficulty]
    I'm not sure how to simplify y. It looks like the fundamental theorem
    of calculus. So e^(-x^2)S(from 0 to x)e^(t^2)dt should just be 1,
    right? I let y=1+c[1]e^(-x^2) and used implicit differentiation, but
    I end up with dy/dx+2xy=2x. If 2x=1 then I'd have it. I also
    substituted y and dy/dx into the equation, but I get the same thing:
    2x=1. I have to be simplifying y wrong. Can you help me? Thanks!

    [Thoughts]
    y=1+ce^(-x^2)
    (y-1)/e^(-x^2)=c
    e^(x^2)y-e^(x^2)=c
    e^(x^2)y'+2xye^(x^2)-2xe^(x^2)=0
    e^(x^2)y'=2xe^(x^2)-2xye^(x^2)
    y'=2x-2xy
    y'+2xy=2x

    dy/dx+2xy=1; y=1+ce^(-x^2); dy/dx=-2xce^(-x^2)
    -2xce^(-x^2)+2x(1+ce^(-x^2))=1
    -2xce^(-x^2)+2x+2xce^(-x^2)=1
    2x=1
    That is difficult to want to go through since it isn't in Latex and is crammed together.

    I would use an integrating factor.

    \displaystyle\int\left\frac{d}{dx}\left(\exp{\left  (\int P(x)dx\right)}y\right)\right]dx=\int\exp{\left(\int P(x)dx\right)}Q(x)dx

    P(x) = 2x

    Q(x) = 1

    What is this y = e^(-x^2)Se^(t^2) dt +c[1]e^(-x^2)

    Another problem?
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  3. #3
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    Quote Originally Posted by dwsmith View Post
    What is this y = e^(-x^2)Se^(t^2) dt +c[1]e^(-x^2)

    Another problem?


    This is the solution, the OP needs the derivative of this function to check if its a solution of the DE given above.
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  4. #4
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    Is it y' = -2xce^(-x^2) ? How can I get Latex? What is an OP?
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  5. #5
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    Quote Originally Posted by duaneg37 View Post
    Is it y' = -2xce^(-x^2) ?
    Maybe does it satisfy \frac{dy}{dx}+2xy = 1

    Quote Originally Posted by duaneg37 View Post
    How can I get Latex?
    Use MATH tags [tex]\frac{dy}{dx}+2xy = 1[/ math] (remove space between "/ math") gives \frac{dy}{dx}+2xy = 1

    Double click on the equation for more info.


    Quote Originally Posted by duaneg37 View Post
    What is an OP?
    Original Post.
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  6. #6
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    You can see how to use LaTeX by looking here. OP stands for "Original Post".

    Your problem: verify the solution

     \displaystyle y = e^{-x^{2}}\int_{0}^{x} e^{t^{2}}\, dt +C\,e^{-x^{2}} for the DE

    y'+2xy=1. Correct?

    You'll need to use the product rule, the chain rule, and the Fundamental Theorem of the Calculus to do this verification:

    \displaystyle y'=\left(e^{-x^{2}}\right)\frac{d}{dx}\left(\int_{0}^{x} e^{t^{2}}\, dt\right)+\left(-2xe^{-x^{2}}\right)\left(\int_{0}^{x} e^{t^{2}}\, dt\right)-2Cxe^{-x^{2}}.

    You'll need to finish this computation, and then plug the result into the DE on the LHS, and see if you get the required 1 to equal the RHS. Make sense?

    [EDIT] Sorry, didn't see pickslides' post.
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  7. #7
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    y'+2xy=1

    \displaystyle y = e^{-x^{2}}\int_{0}^{x} e^{t^{2}}\, dt +C\,e^{-x^{2}} Verify this is a solution of the differential equation above


    \displaystyle y'=\left(e^{-x^{2}}\right)\frac{d}{dx}\left(\int_{0}^{x} e^{t^{2}}\, dt\right)+\left(-2xe^{-x^{2}}\right)\left(\int_{0}^{x} e^{t^{2}}\, dt\right)-2Cxe^{-x^{2}}<br />

    \displaystyle y'=1-2xe^{-x^{2}}\int_{0}^{x}e^{t^{2}}\, dt-2Cxe^{-x^{2}}

    \displaystyle 1-2xe^{-x^{2}}\int_{0}^{x}e^{t^{2}}\, dt-2Cxe^{-x^{2}}+2x\left( e^{-x^{2}}\int_{0}^{x} e^{t^{2}}\, dt +C\,e^{-x^{2}}\right)=1

    \displaystyle 1-2xe^{-x^{2}}\int_{0}^{x}e^{t^{2}}\, dt-2Cxe^{-x^{2}}+2xe^{-x^{2}}\int_{0}^{x}e^{t^{2}}\, dt+2Cxe^{-x^{2}}=1

    \displaystyle 1=1

    I think I've got it now.....Thanks for all your help!
    Last edited by duaneg37; January 18th 2011 at 05:15 PM. Reason: correction
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  8. #8
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    I'm afraid it's incorrect. You can't just get rid of integrals like that. You did correctly use the FTC to simplify the first term, but the second term cannot be simplified. I think you'll find it cancels with terms in the 2xy of the original DE, as does the last term with the constant. Just leave the integrals there!
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