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    Need help with question 3.

    Not entirely sure how to do it do i need to find an integrating factor?
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    (Original post by Super199)
    Need help with question 3.

    Not entirely sure how to do it do i need to find an integrating factor?
    Might be helpful to take a picture of, \LaTeX or link to the question.
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    (Original post by Kvothe the arcane)
    Might be helpful to take a picture of, \LaTeX or link to the question.
    Well i tried uploading it but it wont let me so i shall type it out.

    3. Find the general solution of the differential equation dy/dx-3x^2y= xe^(x^3), giving y explicitly in terms of x. Find also the particular solution for which y=1 when x=0
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    (Original post by Super199)
    Need help with question 3.

    Not entirely sure how to do it do i need to find an integrating factor?
    where is the question
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    (Original post by Super199)
    Well i tried uploading it but it wont let me so i shall type it out.

    3. Find the general solution of the differential equation dy/dx-3x^2y= xe^(x^3), giving y explicitly in terms of x. Find also the particular solution for which y=1 when x=0
    To your question, yes.

    The integrating factor is \displaystyle e^{\int -3x^2 dx}

    Work it out and the equation by it.
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    (Original post by Kvothe the arcane)
    To your question, yes.

    The integrating factor is \displaystyle e^{\int -3x^2 dx}

    Work it out and the equation by it.
    If i could upload a picture it would help but i got the integrating factor as e^-x^3.

    Multiplied through :

    Dy/dxe^-x^3 - 3x^2ye^-x^3= x

    D/dx(ye^-x^3)=x

    Ye^-x^3= x^2/2+c.
    Im having troubles with e. If you divide by e^(-x^3) do you get y= (x^2+c)e^x^3
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    (Original post by Super199)
    If i could upload a picture it would help but i got the integrating factor as e^-x^3.

    Multiplied through :

    Dy/dxe^-x^3 - 3x^2ye^-x^3= x

    D/dx(ye^-x^3)=x

    Ye^-x^3= x^2/2+c.
    Im having troubles with e. If you divide by e^(-x^3) do you get y= (x^2+c)e^x^3
    So you have ye^{-x^3}=\dfrac{x^2}{2}+C
    Perhaps rewrite as \dfrac{y}{e^{x^3}}=\dfrac{x^2}{2  }+C?
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    (Original post by Kvothe the arcane)
    So you have ye^{-x^3}=\dfrac{x^2}{2}+C
    Perhaps rewrite as \dfrac{y}{e^{x^3}}=\dfrac{x^2}{2  }+C?
    Got it cheers need help with another if you dont mind. Im having trouble with the integrating factor again.

    Dy/dx +y/(xlnx)= 1/x. With y=1 when x=e is to be solved using an integrating factor. Find the integrating factor and solve the differential equation. Sketch the solution curve for x>1.
    Integral of xlnx im not sure. Do i have to use parts? But the problem is it is (xlnx)^-1. So not sure. Or can i write it as 1/x*lnx. Idk cheers
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    (Original post by Super199)
    Got it cheers need help with another if you dont mind. Im having trouble with the integrating factor again.

    Dy/dx +y/(xlnx)= 1/x. With y=1 when x=e is to be solved using an integrating factor. Find the integrating factor and solve the differential equation. Sketch the solution curve for x>1.
    Integral of xlnx im not sure. Do i have to use parts? But the problem is it is (xlnx)^-1. So not sure. Or can i write it as 1/x*lnx. Idk cheers
    Had to think about it.

    But it might be helpful to recall that \dfrac{d}{dx}lnx=\dfrac{1}{x} and \displaystyle \int \dfrac{f'(x)}{f(x)}dx=ln(f(x))+C.

    Not IBP, no.
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    (Original post by Kvothe the arcane)
    Had to think about it.

    But it might be helpful to recall that \dfrac{d}{dx}lnx=\dfrac{1}{x} and \displaystyle \int \dfrac{f'(x)}{f(x)}dx=ln(f(x))+C.

    Not IBP, no.
    I cant see the link sorry. I know that lnx integrates to xlnx-x and also what you said. But I cant see the link
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    (Original post by Super199)
    I cant see the link sorry. I know that lnx integrates to xlnx-x and also what you said. But I cant see the link
    \dfrac{1}{xlnx}=\dfrac{1}{x} \times \dfrac{1}{lnx}

    What is the relationship between 1/x and lnx? And how might that be related to what I wrote above?
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    (Original post by Super199)
    Got it cheers need help with another if you dont mind. Im having trouble with the integrating factor again.

    Dy/dx +y/(xlnx)= 1/x. With y=1 when x=e is to be solved using an integrating factor. Find the integrating factor and solve the differential equation. Sketch the solution curve for x>1.
    Integral of xlnx im not sure. Do i have to use parts? But the problem is it is (xlnx)^-1. So not sure. Or can i write it as 1/x*lnx. Idk cheers
    Use the substitution u = \ln x, you'll end up with \int\frac{\mathrm{d}u}{u}.
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    (Original post by Zacken)
    Use the substitution u = \ln x, you'll end up with \int\frac{\mathrm{d}u}{u}.
    Right that helps now how do you integrate lnx/x
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    (Original post by Super199)
    Right that helps now how do you integrate lnx/x
    Not to worry got it. Using parts
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    (Original post by Super199)
    Right that helps now how do you integrate lnx/x
    I'm not sure what you're asking. You want to integrate \int \frac{\mathrm{d}x}{x \ln x} , right? That's \ln (\ln x).

    Integrating \displaystyle \frac{\ln x}{x} is an easy one if you use u = \ln x you should get \int u \, \mathrm{d}u
 
 
 
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