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Thread: differentiation

  1. #1
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    differentiation

    Solve the differential equation

    ydy/dx =x+xy^2 given that y = 0 when x = 0.

    i can get down to here but then i get stuck

    dy/dx= x(1+y^2)/y

    ∫▒y/(1+y^2 ) dy=∫▒ x dx i dont know how to integrate this?
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  2. #2
    Senior Member zzephod's Avatar
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    Re: differentiation

    The differential equation:
    $$
    \frac{dy}{dx}=x \frac{1+y^2}{y}
    $$
    is of variable seperable type reducing to:
    $$
    \int x\;dx=\int\frac{y}{1+y^2}\;dy
    $$
    which integrates to:

    $$
    \frac{x^2}{2}+C=\frac{\ln(1+y^2)}{2}
    $$
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  3. #3
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    Re: differentiation

    How do you integrate the dy
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  4. #4
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    Re: differentiation

    Quote Originally Posted by markosheehan View Post
    How do you integrate the dy
    substitute

    $u=y^2$

    $du=2y~dy$

    you get

    $\dfrac 1 2 \int \dfrac {du}{1+u}$
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  5. #5
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    Re: differentiation

    Quote Originally Posted by romsek View Post
    substitute

    $u=y^2$

    $du=2y~dy$

    you get

    $\dfrac 1 2 \int \dfrac {du}{1+u}$
    A slightly easier substitution would be

    $u=y^2+1$

    $du=2y~dy$

    you get

    $\dfrac 1 2 \int \dfrac {du}{u}$

    Once you have done the integration for $u$ then switch back to $y$
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  6. #6
    Senior Member zzephod's Avatar
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    Re: differentiation

    You recognise it as of the form
    $$
    \frac{1}{2} \int \frac{f'(y)}{f(y)}dy=\frac{1}{2}\int \left( \frac{d}{dy}\ln(f(y)) \right) \;dy=\frac{1}{2} \ln(f(y))
    $$
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  7. #7
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    Re: differentiation

    Quote Originally Posted by markosheehan View Post
    Solve the differential equation

    ydy/dx =x+xy^2 given that y = 0 when x = 0.

    i can get down to here but then i get stuck

    dy/dx= x(1+y^2)/y

    ∫▒y/(1+y^2 ) dy=∫▒ x dx i dont know how to integrate this?
    It can easily be solved by using integration by substitution
    $\int {\frac{y}{1+y^{2}}dy}= \int{x dx}$
    Substitute $1+y^{2} = u$
    $\implies{ydy = \frac{du}{2}}$
    $\int{\frac{du}{2u}}= \int{x dx}$
    $\frac{1}{2}lnu=\frac{x^{2}}{2}+C$
    Put the value of u
    $\frac{1}{2}ln(1+y^{2})=\frac{x^{2}}{2}+C$
    For more practice of integartion by substitution http://www.actucation.com/calculus-2...y-substitution
    Last edited by rahuljoshi; Oct 24th 2016 at 03:42 AM.
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