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    I'm having some trouble with getting the general solutions of these differential equations! They should be pretty easy, they're at the start of the misc ex

    x \frac{dy}{dx} + (x+1)y = 1

    and for -1<x<1

    (1 + x^2)\frac{dy}{dx} -xy +1 = 0

    If someone could post a method for one of them, it would be super helpful! Thank you!
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    (Original post by kawehi)
    I'm having some trouble with getting the general solutions of these differential equations! They should be pretty easy, they're at the start of the misc ex

    x \frac{dy}{dx} + (x+1)y = 1

    and for -1<x<1

    (1 + x^2)\frac{dy}{dx} -xy +1 = 0

    If someone could post a method for one of them, it would be super helpful! Thank you!
    Do you know the general form for solving 1st ODEs?
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    (Original post by kawehi)
    I'm having some trouble with getting the general solutions of these differential equations! They should be pretty easy, they're at the start of the misc ex

    x \frac{dy}{dx} + (x+1)y = 1

    and for -1<x<1

    (1 + x^2)\frac{dy}{dx} -xy +1 = 0

    If someone could post a method for one of them, it would be super helpful! Thank you!
    start by dividing the coefficient of dy/dx
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    (Original post by kawehi)
    I'm having some trouble with getting the general solutions of these differential equations! They should be pretty easy, they're at the start of the misc ex

    x \frac{dy}{dx} + (x+1)y = 1

    and for -1<x<1

    (1 + x^2)\frac{dy}{dx} -xy +1 = 0

    If someone could post a method for one of them, it would be super helpful! Thank you!
    Have you heard of integrating factors?
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    (Original post by kawehi)
    ...
    If this is a 1st ODE,

    \displaystyle \frac{\mathrm{d} y}{\mathrm{d}x} + Py = Q \ \text{(I)}

    Then the general solution can be written as

    \displaystyle \text{General solution:} \ e^{\int P \mathrm{d}x}y = \int e^{\int P\mathrm{d}x}Q \ \mathrm{d}x

    where e^{\int P \mathrm{d}x} is the integrating factor. (Click on the annotated video to see the proof!)

    Once you've got it in the required form as in (I) (which you can do using TeeEm's hint), you should find the integrating factor and then the general solution should be a cinch to find from there!
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    (Original post by aymanzayedmannan)
    \int e^{P\mathrm{d}x}Q \ \mathrm{d}x
    Teensy typo. :-)
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    (Original post by aymanzayedmannan)
    Do you know the general form for solving 1st ODEs?
    (Original post by TeeEm)
    start by dividing the coefficient of dy/dx
    Yeah, I know the general method but I can't seem to get the 'product rule' bit..

    

x  (dy/dx) + (x+1)y = 1

(dy/dx) + (1 + 1/x)y = 1/x



I(x) = \int (1 + \frac{1}{x})dx = x + ln x

So multiply the original thing by xe^x



xe^x (\frac{dy}{dx} + (1 + \frac{1}{x})y = \frac{1}{x})

xe^x \frac{dy}{dx} + xe^x (1 + \frac{1}{x})y = e^x

    Sorry the latex fracs are a bit dodgy haha, but I don't know what to do from there.. Have I made any mistakes or can I just not spot the product rule bit?
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    (Original post by kawehi)
    Yeah, I know the general method but I can't seem to get the 'product rule' bit..

    

x  (dy/dx) + (x+1)y = 1

(dy/dx) + (1 + 1/x)y = 1/x



I(x) = \int (1 + \frac{1}{x})dx = x + ln x

So multiply the original thing by xe^x



xe^x (\frac{dy}{dx} + (1 + \frac{1}{x})y = \frac{1}{x})

xe^x \frac{dy}{dx} + xe^x (1 + \frac{1}{x})y = e^x

    Sorry the latex fracs are a bit dodgy haha, but I don't know what to do from there.. Have I made any mistakes or can I just not spot the product rule bit?
    I let aymanzayedmannan continue as he was first and I found myself tied up now.
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    (Original post by TeeEm)
    I let aymanzayedmannan continue as he was first and I found myself tied up now.
    Ok, thanks
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    (Original post by kawehi)
    Yeah, I know the general method but I can't seem to get the 'product rule' bit..

    

xe^x \frac{dy}{dx} + xe^x (1 + \frac{1}{x})y = e^x

    Sorry the latex fracs are a bit dodgy haha, but I don't know what to do from there.. Have I made any mistakes or can I just not spot the product rule bit?
    From there: the LHS is nothing but \displaystyle \frac{d}{dx} \left(xe^x y\right) = e^x

    Now you can integrate both sides...
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    (Original post by Zacken)
    From there: the LHS is nothing but \displaystyle \frac{d}{dx} \left(xe^x y\right) = e^x

    Now you can integrate both sides...
    Ohh, I somehow missed that you had to use product rule on xe^x :rolleyes: Thanks!!!
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    (Original post by kawehi)
    Ohh, I somehow missed that you had to use product rule on xe^x :rolleyes: Thanks!!!
    The whole point of using the integrating factor is that you get your DE in the form:

    \displaystyle \frac{d}{dx} \left(Iy\right) = Iq(x) so you can integrate both sides and solve. :-)
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    (Original post by Zacken)
    Teensy typo. :-)
    Still more helpful than you.
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    (Original post by kawehi)
    Yeah, I know the general method but I can't seem to get the 'product rule' bit..

    

x  (dy/dx) + (x+1)y = 1

(dy/dx) + (1 + 1/x)y = 1/x



I(x) = \int (1 + \frac{1}{x})dx = x + ln x

So multiply the original thing by xe^x



xe^x (\frac{dy}{dx} + (1 + \frac{1}{x})y = \frac{1}{x})

xe^x \frac{dy}{dx} + xe^x (1 + \frac{1}{x})y = e^x

    Sorry the latex fracs are a bit dodgy haha, but I don't know what to do from there.. Have I made any mistakes or can I just not spot the product rule bit?
    Alright, so you're gold on finding the integrating factor :borat: You're struggling on the "multiplying through" part from what it seems.

    Now, after you've divided through by x, your DE is now \displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} + \left ( 1 +\frac{1}{x} \right )y = \frac{1}{x}, right? Comparing this to the equation (I) I gave you, you can see that Q = 1/x.

    Rather than multiplying it out by your factor, try using the general solution I gave you. You will see that you'll get an easily integrable function by multiplying through using that method.

    EDIT: looks like Zacken took care of it while I was struggling with Latex!

    (Original post by Zacken)
    Teensy typo. :-)
    thank you
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    (Original post by tinkerbella~)
    Still more helpful than you.
    not at all
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    (Original post by aymanzayedmannan)
    Spoiler:
    Show
    It will lead to \displaystyle y(x+\ln x) = \int \frac{x + lnx}{x} \mathrm{d}x If I'm not doing this incorrectly!



    thank you
    I think the integrating factor is e^{x + \ln x} = xe^x. The user has just done some abuse of notation. :-)
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    (Original post by tinkerbella~)
    Still more helpful than you.
    Play nice.
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    (Original post by Zacken)
    I think the integrating factor is e^{x + \ln x} = xe^x. The user has just done some abuse of notation. :-)
    nah, mate I realise and edited it because it looked iffy.
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    (Original post by Zacken)
    Play nice.
    That's not my style.
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    (Original post by aymanzayedmannan)
    nah, mate I realise and edited it because it looked iffy.
    Go write some music, enough maths. :rock:

    (Original post by tinkerbella~)
    That's not my style.
    You don't have style.
 
 
 
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