Results 1 to 3 of 3

Thread: dy/dx = -y/x

  1. #1
    Newbie
    Joined
    Jul 2009
    Posts
    1

    dy/dx = -y/x

    Hello everybody,

    I need to solve

    dy/dx = -y/x

    for some economics problem. Can anybody help me?

    Thanks,

    Rob
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    6
    The DE you propose can be written as...

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

    Al thet you have to do is to integrate first and second term of (1) indicating the 'arbitrary constant' as $\displaystyle \ln c$...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Quote Originally Posted by robvaneijk View Post
    Hello everybody,

    I need to solve

    dy/dx = -y/x

    for some economics problem. Can anybody help me?

    Thanks,

    Rob
    $\displaystyle \frac{dy}{dx} = -\frac{y}{x}$

    $\displaystyle \frac{1}{y}\,\frac{dy}{dx} = -\frac{1}{x}$

    $\displaystyle \int{\frac{1}{y}\,\frac{dy}{dx}\,dx} = \int{-\frac{1}{x}\,dx}$

    $\displaystyle \int{\frac{1}{y}\,dy} = -\int{\frac{1}{x}\,dx}$

    $\displaystyle \ln{|y|} + C_1 = -\ln{|x|} + C_2$

    $\displaystyle \ln{|y|} + \ln{|x|} = C$, where $\displaystyle C = C_2 - C_1$

    $\displaystyle \ln{|y||x|} = C$

    $\displaystyle \ln{|xy|} = C$

    $\displaystyle |xy| = e^C$

    $\displaystyle xy = \pm e^C$

    $\displaystyle xy = A$ where $\displaystyle A = \pm e^C$

    $\displaystyle y = \frac{A}{x}$.
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum