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Math Help - can anyone help me out solving differential equa.?? plz

  1. #1
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    can anyone help me out solving differential equa.?? plz

    can anyone help me out wif this problem??
    maybe...its an easy cake for u..

    dy/dx=xe^(y-x^2) and y(0)=0

    I have to solve the initial value....

    and show all intermediate steps in my solution of the differential equation.

    thanks
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  2. #2
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    Quote Originally Posted by mookie View Post
    can anyone help me out wif this problem??
    maybe...its an easy cake for u..

    dy/dx=xe^(y-x^2) and y(0)=0

    I have to solve the initial value....

    and show all intermediate steps in my solution of the differential equation.

    thanks
    The ODE separates

     <br />
\int e^{-y}\,dy = \int x e^{-x^2}\,dx + c<br />
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  3. #3
    MHF Contributor Calculus26's Avatar
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    rewrite

    dy/dx = x e^y e^(-x^2)

    Separate

    e^(-y) dy = x e^(-x^2) dx

    Now you have 2 simple integrals so integrate

    Finally apply the initial condition to determine C
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  4. #4
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    integration..

    can u plz how me nxt steps??
    integrations

    Quote Originally Posted by danny arrigo View Post
    The ODE separates

     <br />
\int e^{-y}\,dy = \int x e^{-x^2}\,dx + c<br />
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  5. #5
    MHF Contributor Calculus26's Avatar
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    The LHS is simply -e^(-y)

    for the RHS use the u substitution u = -x^2


    you obtain -e^(-y) = -1/2 e^(-x^2) + C

    e^(-y) = 1/2 e^(-x^2) + C

    since y(0) = 0

    1 = 1/2 + C C =1/2

    e^(-y) = 1/2 e^(-x^2) + 1/2

    now use logarithms to solve for y-- I/ll let you finish that part
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  6. #6
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    from here...

    int(e^-y)dy=int(xe^(-x^2))dx,

    Let u=-x^2
    du/dx=-2x
    dx=du/-2x
    then, it becomes,

    int(e^-y)dy=int(-e^u)du
    -------------------------
    this is the part that confuses me!!!

    would it become...

    -e^-y= -e^u +C
    -e^-y=-e^(-x^2) +C

    C= -e^-y + e^(-x^2)

    given that, y(0)=0,
    substitute y=0 and x=0 to find C,

    then,

    C=-1 + 1=0

    is this right???




    Quote Originally Posted by Calculus26 View Post
    rewrite

    dy/dx = x e^y e^(-x^2)

    Separate

    e^(-y) dy = x e^(-x^2) dx

    Now you have 2 simple integrals so integrate

    Finally apply the initial condition to determine C
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  7. #7
    MHF Contributor Calculus26's Avatar
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    not quite

    int(e^-y)dy=int(-e^u)du

    should be

    int(e^-y)dy=-1/2int(-e^u)du

    remember -du/2 = xdx
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  8. #8
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    thx!!!so much

    so, i got

    -e^(-y)= 1/2 e^(-x^2)+1/2...

    how do u use logarithm to find out y

    lol...i m just so dumb...


    Quote Originally Posted by Calculus26 View Post
    The LHS is simply -e^(-y)

    for the RHS use the u substitution u = -x^2


    you obtain -e^(-y) = -1/2 e^(-x^2) + C

    e^(-y) = 1/2 e^(-x^2) + C

    since y(0) = 0

    1 = 1/2 + C C =1/2

    e^(-y) = 1/2 e^(-x^2) + 1/2

    now use logarithms to solve for y-- I/ll let you finish that part
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  9. #9
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    I get all your explanations thanks a lot!!!

    but.. shouldn't C be equal to -1/2

    -e^(-y) = -1/2 e^(-x^2) + C

    since, y(0)=0
    then,
    doesn't it become:

    -1=-1/2 +C
    C= -1/2

    Thus, the expression is now then,
    -e^(-y) = -1/2 e^(-x^2) -1/2????

    and also not sure how to solve y----.. T_T using logarithm

    thanks you so much

    Quote Originally Posted by Calculus26 View Post
    The LHS is simply -e^(-y)

    for the RHS use the u substitution u = -x^2


    you obtain -e^(-y) = -1/2 e^(-x^2) + C

    e^(-y) = 1/2 e^(-x^2) + C

    since y(0) = 0

    1 = 1/2 + C C =1/2

    e^(-y) = 1/2 e^(-x^2) + 1/2

    now use logarithms to solve for y-- I/ll let you finish that part
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  10. #10
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    using logarithm..to solve y..need help!! thx

    i have to solve y---, solving the initial value

    given, dy/dx=xe^(y-x^2) and y(0)=0

    using differential equation,

    I hav done it so far up to this step...

    ------------------------------

    -e^(-y) = -1/2 e^(-x^2) + C

    since, y(0)=0
    then,
    doesn't it become:

    -1=-1/2 +C
    C= -1/2

    Thus, the expression is now then,
    -e^(-y) = -1/2 e^(-x^2) -1/2????

    and also not sure how to solve y----.. T_T using logarithm for the next step..

    thanks you so muc
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  11. #11
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    Quote Originally Posted by mookie View Post
    i have to solve y---, solving the initial value

    given, dy/dx=xe^(y-x^2) and y(0)=0

    using differential equation,

    I hav done it so far up to this step...

    ------------------------------

    -e^(-y) = -1/2 e^(-x^2) + C

    since, y(0)=0
    then,
    doesn't it become:

    -1=-1/2 +C
    C= -1/2

    Thus, the expression is now then,
    -e^(-y) = -1/2 e^(-x^2) -1/2????

    and also not sure how to solve y----.. T_T using logarithm for the next step..

    thanks you so muc
    -e^{-y}=-\frac{1}{2}e^{-x^2}+c

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

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

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

    Now using the IC

    0=\ln(\frac{1}{2}e^{0}+c) \iff 0=\ln(\frac{1}{2}+c) \iff 1=\frac{1}{2}+c \iff \frac{1}{2}=c
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