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Math Help - Differential Equations help please!

  1. #1
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    Differential Equations help please!

    hi, im having problems with these...

    find the general solutions for:

    t^2 dx/dt = x +1


    and

    y^1/2 dy/dx = y/x


    if someone could go step by step on those i would be very happy!
    please help!

    many thanks
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by united_legend View Post
    hi, im having problems with these...

    find the general solutions for:

    t^2 dx/dt = x +1


    and

    y^1/2 dy/dx = y/x


    if someone could go step by step on those i would be very happy!
    please help!

    many thanks
    Theser are SDEs(seperable differential equations)...for example the first one t^2\frac{dx}{dt}=x+1...sepearing we get \frac{1}{x+1}dx=\frac{dt}{t^2}...integrating both sides we get ln|x+1|=\frac{-1}{t}+C now solving for the function x which is what we want we get x=Ce^{\frac{-1}{t}}-1 where C is a constant..
    Last edited by Mathstud28; April 20th 2008 at 08:53 AM.
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by united_legend View Post
    hi, im having problems with these...

    find the general solutions for:

    t^2 dx/dt = x +1


    and

    y^1/2 dy/dx = y/x


    if someone could go step by step on those i would be very happy!
    please help!

    many thanks
    Now you should submit your work to see if you are doing it correctly but the answer to 2 is y=\bigg(\frac{\ln(x)}{2}+C\bigg)^2
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  4. #4
    Moo
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    Hellooo,

    Quote Originally Posted by Mathstud28 View Post
    integrating both sides we get [tex]ln|x+1|=\frac{t^3}{3}+C[/math
    Ow, really ?

    I thought that 1/tē was the derivative for -1/t
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  5. #5
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    i still cant do it... :/

    sorry if i sound completely stupid, but how do you seperate it to get that?

    cheers!
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    Hellooo,



    Ow, really ?

    I thought that 1/tē was the derivative for -1/t
    Ehhhhhhhhhhh.......I have gotten to the point where I dont care if you mock me because I think you know it isnt because I dont understand
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by united_legend View Post
    i still cant do it... :/

    sorry if i sound completely stupid, but how do you seperate it to get that?

    cheers!
    You want to get the dx's with the x's and the dt's with the dt.....and then integrate as you normally would then solve for teh function they ask you to(the function you are supposed to solve for is the whatever letter is on the top of the fraction...example \frac{dy}{dx} it would be solve for y)
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  8. #8
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    t^2 dx/dt = x+1

    integrate 1/x+1 dx = integrate t^-2 dt

    ?

    which is ln x+1 = -t^-1


    :S wat am i doing wrong?
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  9. #9
    Moo
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    No, this is correct

    Don't forget the constant of integration !

    (moo)
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by united_legend View Post
    t^2 dx/dt = x+1

    integrate 1/x+1 dx = integrate t^-2 dt

    ?

    which is ln x+1 = -t^-1


    :S wat am i doing wrong?
    Yeah if you will notice I accidentally integrated t^2 instead of \frac{1}{t^2}...but as you will also notice...I corrected myeslef awhile ago...you are doing it correctly
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  11. #11
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    Oh right, thanks

    and for the 2nd one i get..

    y^-2 dy/dx = y/x

    integrate y^1/2 /y dy = integrate 1/x dx

    integrate y^-1/2 dy = integrate x^-1 dx

    = 1/2 y^1/2 = x

    ?

    i get confused with the last bit, you add 1 to the power and divide by the new power dont you?

    thanks a lot so far!
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  12. #12
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by united_legend View Post
    Oh right, thanks

    and for the 2nd one i get..

    y^-2 dy/dx = y/x

    integrate y^1/2 /y dy = integrate 1/x dx

    integrate y^-1/2 dy = integrate x^-1 dx

    = 1/2 y^1/2 = x

    ?

    i get confused with the last bit, you add 1 to the power and divide by the new power dont you?

    thanks a lot so far!
    you have \sqrt{y}\frac{dy}{dx}=\frac{y}{x} dividing both sides by y we get y^{\frac{-1}{2}}dy=dx\frac{1}{x}..integrating we get 2\sqrt{y}=ln|x|+c make sense from there?
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  13. #13
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    Quote Originally Posted by Mathstud28 View Post
    you have \sqrt{y}\frac{dy}{dx}=\frac{y}{x} dividing both sides by y we get y^{\frac{-1}{2}}dy=dx\frac{1}{x}..integrating we get 2\sqrt{y}=ln|x|+c make sense from there?
    i dont no how you get the 2\sqrt{y} bit

    cheers
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  14. #14
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by united_legend View Post
    i dont no how you get the 2\sqrt{y} bit

    cheers
    \int{y^{\frac{-1}{2}}dy}=\frac{y^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}=2y^{\frac{1}{2}}=2\sqrt{y}
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