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    Find the particular solution:

    dy/dx=x(y+1) .....y=0 when x=1

    This is what I did:

    integrated 1/y+1 dy = integrated x dx

    y+ ln (y)=(1/2)x^2+c
    e^y+y =e^(1/2x^2) + c

    but my answer was wrong!
    I don't know when to substitute the values in because lny ...ln (0) does not work
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    It should be

     ln(y + 1) = \frac {1}{2} x^2 + c
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    (Original post by 0utdoorz)
    Find the particular solution:

    dy/dx=x(y+1) .....y=0 when x=1

    This is what I did:

    integrated 1/y+1 dy = integrated x dx

    y+ ln (y)=(1/2)x^2+c
    e^y+y =e^(1/2x^2) + c

    but my answer was wrong!
    I don't know when to substitute the values in because lny ...ln (0) does not work
    how did you get that?
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    (Original post by 0utdoorz)
    Find the particular solution:

    dy/dx=x(y+1) .....y=0 when x=1

    This is what I did:

    integrated 1/y+1 dy = integrated x dx

    y+ ln (y)=(1/2)x^2+c
    e^y+y =e^(1/2x^2) + c

    but my answer was wrong!
    I don't know when to substitute the values in because lny ...ln (0) does not work
    you seem to think that 1/(y+1) = (1/y) + 1, which it doesn't
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    I've done ln y+1 = 1/2x^2+c

    substituted in x=1 y=0

    ln 1 = 1/2 +c
    0 = 1/2 +c
    c=-1/2

    but c is meant to be -1?


    also
    on the RHS I got 1/2x^2 +c

    but the answer is meant to be y=e^((x^2-1)/2) -1
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    (Original post by 0utdoorz)
    I've done ln y+1 = 1/2x^2+c

    substituted in x=1 y=0

    ln 1 = 1/2 +c
    0 = 1/2 +c
    c=-1/2

    but c is meant to be -1?

    also
    on the RHS I got 1/2x^2 +c

    but the answer is meant to be y=e^((x^2-1)/2) -1
    Hmmm.

     \frac {dy}{dx} = xy - x

     -xy dy = -x dx

     \int y dy =  \int 1 dx

     \frac {y^2}{2} = x + c

     0 = 1 + c

     -1 = c

    ?
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    (Original post by 0x2a)
    Hmmm.

     \frac {dy}{dx} = xy - x

     -xy dy = -x dx

     \int y dy =  \int 1 dx

     \frac {y^2}{2} = x + c

     0 = 1 + c

     -1 = c

    ?
    thanks!! but how do you get the e^((x^2-1)/2) part?
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    (Original post by 0utdoorz)
    thanks!! but how do you get the e^((x^2-1)/2) part?

    \displaystyle \int \dfrac{1}{1+y} \ dy = \ln(1+y) + C \neq \ln y + y +C

    Once you've established that, go about your business in the normal way and it should turn out (though I haven't verified the answer for myself).
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    (Original post by 0utdoorz)
    thanks!! but how do you get the e^((x^2-1)/2) part?
    It should be  e^{\frac{x^2}{2}} - 1 + c I think.....

    You should remind yourself what  ln x is, then it'll all become clear to you

    Spoiler:
    Show
     ln(y + 1) = \frac {x^2}{2} + c which is equivalent to  y + 1 = e^{\frac{x^2}{2}} + c
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    (Original post by 0x2a)
    It should be  e^{\frac{x^2}{2}} - 1 + c I think.....

    You should remind yourself what  ln x is, then it'll all become clear to you

    Spoiler:
    Show
     ln(y + 1) = \frac {x^2}{2} + c which is equivalent to  y + 1 = e^{\frac{x^2}{2}} + c
    thank you!
    from there, how do you get c=-1? (where y=0 x=1)

    because when I do the x=1 into e^(x^2/2) it doesn't work out?
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    (Original post by 0utdoorz)
    thank you!
    from there, how do you get c=-1? (where y=0 x=1)
    You have the given values so to find C just sub in y= 0 and x = 1 and you will find C
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    I will go through this thread and piece it all together.

    Initial conditions y = 0 , x = 1

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

    which is separable

    \displaystyle \frac{1}{y+1} \ dy = x \ dx

    \displaystyle \int \frac{1}{y+1} \ dy = \int x \ dx

    \displaystyle \ln(y+1) = \frac{x^2}{2} + C_1

    Hang on, couldn't you just sub in here to find C ? And then take the exp after?
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    (Original post by SubAtomic)
    I will go through this thread and piece it all together.

    Initial conditions y = 0 , x = 1

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

    which is separable

    \displaystyle \frac{1}{y+1} \ dy = x \ dx

    \displaystyle \int \frac{1}{y+1} \ dy = \int x \ dx

    \displaystyle \ln(y+1) = \frac{x^2}{2} + C_1

    \displaystyle y + 1 = e^{\frac{x^2}{2}} + C

    Now sub your vals in and rearrange.
    when I substitute the values in, I get 1=e^(1/2) +c

    1-e^(1/2) doesn't equal -1(the answer)?

    or if I log it
    I get ln 1=1/2 +c
    c=-1/2?
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    (Original post by 0utdoorz)
    thank you!
    from there, how do you get c=-1? (where y=0 x=1)

    because when I do the x=1 into e^(x^2/2) it doesn't work out?
    I think I've made a mistake...

     ln(y + 1) = \frac {x^2}{2} + c

     y + 1 = e^{\frac {x^2}{2} + c}

    1 = e^{\frac{1}{2} + c}

    In which case c must be  - \frac{1}{2}

     c = - \frac{1}{2}

     c = -1

    .....
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    (Original post by 0utdoorz)
    when I substitute the values in, I get 1=e^(1/2) +c

    1-e^(1/2) doesn't equal -1(the answer)?

    or if I log it
    I get ln 1=1/2 +c
    c=-1/2?
    Ignore that post I edited while you were typing it should be Ae^{stuff} - 1 , be careful when you start playing with things once they have been integrated, C isn't C if you exp it.

    You know what ln(1) is ?
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    (Original post by SubAtomic)
    I will go through this thread and piece it all together.

    Initial conditions y = 0 , x = 1

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

    which is separable

    \displaystyle \frac{1}{y+1} \ dy = x \ dx

    \displaystyle \int \frac{1}{y+1} \ dy = \int x \ dx

    \displaystyle \ln(y+1) = \frac{x^2}{2} + C_1

    Hang on, couldn't you just sub in here to find C ? And then take the exp after?
    Your post has made everything I learned for my A Levels come flooding back. How very sad that I genuinely enjoyed A Level maths
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    (Original post by NessEB)
    Your post has made everything I learned for my A Levels come flooding back. How very sad that I genuinely enjoyed A Level maths
    Indeed. The syllabus is disappointingly simplistic (although this is easy to say when you're doing uni maths ).
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    (Original post by NessEB)
    Your post has made everything I learned for my A Levels come flooding back. How very sad that I genuinely enjoyed A Level maths
    That is reassuring, it means I am not completely pathetic at A-level maths hahaha
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    a. Separate the variables

    b. Integrate your equation from part a to find the general solution of the differential equation

    c. Use the intial conditions to determine the value of the constants of integration.

    d. State the particular solution of the differential equation using your constants from part c.
 
 
 
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