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Math Help - not a perfect derivative

  1. #1
    Member i_zz_y_ill's Avatar
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    not a perfect derivative

    Ive come across this derivative question:
     (y+xexp(x^2))\frac{dy}{dx}+x+(1+2x^2)yexp(x^2)=0
    The solution is quite simple up until the solution, basically I understand that \frac{d}{dx}(y+xexp(x^2))=exp(x^2)+2x^2exp(x^2)=ex  p(x^2)(1+2x^2) and that also \frac{d}{dy}(x+(1+2x^2)yexp(x^2))=exp(x^2)(1+2x^2) as shown is the solutions but I fail to understand how the answer: \frac{x^2+y^2}{2}+xyexp(x^2)=C is derived from this point. I have tried integratind what would be the right hand side to the equivalent variable on the LHS and all fails. Im afraid that im only used to perfect derivatives when you can spot the product rule. Any suggestions on th method in such case or how to proceed?
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  2. #2
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    Quote Originally Posted by i_zz_y_ill View Post
    Ive come across this derivative question:
     (y+xexp(x^2))\frac{dy}{dx}+x+(1+2x^2)yexp(x^2)=0
    The solution is quite simple up until the solution, basically I understand that \frac{d}{dx}(y+xexp(x^2))=exp(x^2)+2x^2exp(x^2)=ex  p(x^2)(1+2x^2) and that also \frac{d}{dy}(x+(1+2x^2)yexp(x^2))=exp(x^2)(1+2x^2) as shown is the solutions but I fail to understand how the answer: \frac{x^2+y^2}{2}+xyexp(x^2)=C is derived from this point. I have tried integratind what would be the right hand side to the equivalent variable on the LHS and all fails. Im afraid that im only used to perfect derivatives when you can spot the product rule. Any suggestions on th method in such case or how to proceed?
    If I understand you correctly I think you are talking about an Exact differential

    This would be a function

    z(x,y) such that

    dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}=0

    To recover z we would use partial integration.

    so in your example

     (y+xe^{x^2})\frac{dy}{dx}+x+(1+2x^2)ye^{x^2}=0

    Now multiplying by dx we get

     (y+xe^{x^2})dy+[x+(1+2x^2)ye^{x^2}]dx=0

    Now lets take the first part

    (y+xe^{x^2})dy This is the partial of z with respect to y

    \frac{\partial z}{\partial y}= (y+xe^{x^2})dy

    Now if we integrate with respect to y we get

    z(x,y)=\frac{1}{2}y^2+xye^{x^2}+g(x)

    where the arbitrary constant is a function of x.

    Now we know what the partial of z with respect to x should be so

    \frac{\partial z}{\partial x}=ye^{x^2}+2x^2ye^{x^2}+g'(x)=(1+2x^2)ye^{x^2}+g'  (x)

    Now we set this equal to the part above to get

    (1+2x^2)ye^{x^2}+g'(x)=[x+(1+2x^2)ye^{x^2}]

    Simplifying we get

    g'(x)=x \iff g(x)=\frac{1}{2}x^2

    Now putting this back into z(x,y) we get

    \frac{1}{2}y^2+xye^{x^2}+\frac{1}{2}x^2=C
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  3. #3
    Member i_zz_y_ill's Avatar
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    thanks but is it not dz=\frac{dz}{dx}dx+\frac{dz}{dy}dy=0

    not dz=\frac{dz}{dx}dx+\frac{dz}{dy}=0?
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  4. #4
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    Quote Originally Posted by i_zz_y_ill View Post
    thanks but is it not dz=\frac{dz}{dx}dx+\frac{dz}{dy}dy=0

    not dz=\frac{dz}{dx}dx+\frac{dz}{dy}=0?
    ummm...... Yes it is, and my above answer is the same as the answer you posted (just rearrange and factor).

    Here is what you started with

    (y+xexp(x^2))\frac{dy}{dx}+x+(1+2x^2)yexp(x^2)=0

    Now multiply the whole equation by dx (Trust me)

    Now you get

    (y+xexp(x^2))dy+[x+(1+2x^2)yexp(x^2)]dx=0


    okay now we have

    \underbrace{(y+xexp(x^2))}_{\frac{\partial z}{\partial y}}dy+\underbrace{[x+(1+2x^2)yexp(x^2)]}_{\frac{\partial z }{\partial x}}dx=0

    So this is in the form

    \frac{\partial z}{\partial y}dy+\frac{\partial z}{\partial x}dx=0

    This is the form I described in my post above

    Exact differential - Wikipedia, the free encyclopedia
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  5. #5
    Member i_zz_y_ill's Avatar
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    ok but why dont you properly integrate x+(1+2x^2)yexp(x^2) like you did for \frac{dz}{dx}''.
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  6. #6
    Member i_zz_y_ill's Avatar
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    erm i got \frac{x^2+y^2}{2}+exp(x^2)((y^2\frac{1}{2}+x^2)+xy  )

    ok im just gonna accept hat you said in the first post that i know what the partial of dz/dx is because before we mutiplied through by dx we already had it on its own whilst for dz/dy we didnt, purely on that basis,,do you happen to know how to solve the d.e using y=x^m on my other thread dude?
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