Results 1 to 9 of 9

Math Help - Differential equation with substitution

  1. #1
    Senior Member
    Joined
    Apr 2010
    Posts
    487

    Differential equation with substitution

    The question:

    Use the substitution v = \frac{y}{x} to solve

    \frac{dy}{dx} = \frac{(\frac{y}{x})^3 + 1}{(\frac{y}{x})^2}

    given that y = 1 at x = 1.

    My attempt:
    \frac{dy}{dx} = \frac{v^3 + 1}{v^2}
    y = \int{\frac{v^3}{v^2}} + \int{\frac{1}{v^2}}
    = \int{v} + \int{\frac{1}{v^2}}
    = \frac{v^2}{2} - \frac{1}{v} + C
    = \frac{1}{2}(\frac{y}{x})^2} - \frac{1}{\frac{y}{x}} + C
    = \frac{1}{2}(\frac{y}{x})^2} - \frac{x}{y} + C

    Substituting initial value gives C = 3/2

    Is this correct? I don't have answers for practice exams.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    \displaystyle f(1,1)=\frac{1}{2}*1^2-1+c\Rightarrow c=\frac{1}{2}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Glitch View Post
    The question:

    Use the substitution v = \frac{y}{x} to solve

    \frac{dy}{dx} = \frac{(\frac{y}{x})^3 + 1}{(\frac{y}{x})^2}

    given that y = 1 at x = 1.

    My attempt:
    \frac{dy}{dx} = \frac{v^3 + 1}{v^2}
    y = \int{\frac{v^3}{v^2}} + \int{\frac{1}{v^2}}
    = \int{v} + \int{\frac{1}{v^2}}
    = \frac{v^2}{2} - \frac{1}{v} + C
    = \frac{1}{2}(\frac{y}{x})^2} - \frac{1}{\frac{y}{x}} + C
    = \frac{1}{2}(\frac{y}{x})^2} - \frac{x}{y} + C

    Substituting initial value gives C = 3/2

    Is this correct? I don't have answers for practice exams.
    No, that's not correct. You tried to solve the problem with 3 variables. You integrated both sides with respect to x while treating v as the variable too. absurdity

    \displaystyle v = \frac yx \implies y = vx

    \displaystyle \Rightarrow \frac {dy}{dx} = \frac {dv}{dx} \cdot x + v

    You must substitute that for dy/dx into your differential equation then proceed via separation of variables.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,404
    Thanks
    1293
    No, it's not. You need to get a DE that does not have any reference to the variable \displaystyle y.


    \displaystyle \frac{dy}{dx} = \frac{\left(\frac{y}{x}\right)^3 + 1}{\left(\frac{y}{x}\right)^2}.

    Make the substitution \displaystyle v = \frac{y}{x}. Then \displaystyle y = v\,x and \displaystyle \frac{dy}{dx} = x\,\frac{dv}{dx} + v and the DE becomes

    \displaystyle x\,\frac{dv}{dx} + v = \frac{v^3 + 1}{v^2}

    \displaystyle x\,\frac{dv}{dx} + v = v + \frac{1}{v^2}

    \displaystyle x\,\frac{dv}{dx} = \frac{1}{v^2}

    \displaystyle v^2\,\frac{dv}{dx} = \frac{1}{x}

    \displaystyle \int{v^2\,\frac{dv}{dx}\,dx} = \int{\frac{1}{x}\,dx}

    \displaystyle \int{v^2\,dv} = \ln{|x|} + C_1

    \displaystyle \frac{v^3}{3} + C_2 = \ln{|x|} + C_1

    \displaystyle \frac{v^3}{3} = \ln{|x|} + C where \displaystyle C = C_1-C_2

    \displaystyle v^3 = 3\ln{|x|} + 3C

    \displaystyle v = \sqrt[3]{3\ln{|x|} + 3C}

    \displaystyle \frac{y}{x} = \sqrt[3]{3\ln{|x|} + 3C}

    \displaystyle y = x\sqrt[3]{3\ln{|x|} + 3C}.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
     \displaystyle y = vx which implies  \displaystyle \frac{dy}{dx} = x\frac{dv}{dx} + v by the chain rule (and product rule).
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Santa Cruz, CA
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by Glitch View Post
    The question:

    Use the substitution v = \frac{y}{x} to solve

    \frac{dy}{dx} = \frac{(\frac{y}{x})^3 + 1}{(\frac{y}{x})^2}

    given that y = 1 at x = 1.

    My attempt:
    \frac{dy}{dx} = \frac{v^3 + 1}{v^2}
    y = \int{\frac{v^3}{v^2}} + \int{\frac{1}{v^2}}
    = \int{v} + \int{\frac{1}{v^2}}
    = \frac{v^2}{2} - \frac{1}{v} + C
    = \frac{1}{2}(\frac{y}{x})^2} - \frac{1}{\frac{y}{x}} + C
    = \frac{1}{2}(\frac{y}{x})^2} - \frac{x}{y} + C

    Substituting initial value gives C = 3/2

    Is this correct? I don't have answers for practice exams.
    I'm sorry to say that this is not correct, because you're integrating with respect to the wrong variable (v instead of x).

    First apply the substitution that's given to you. v=\frac{y}{x}. So now we need to rewrite \dfrac{\,dy}{\,dx}:

    \dfrac{\,dv}{\,dx}=-\dfrac{y}{x^2}+\dfrac{1}{x}\dfrac{\,dy}{\,dx}. Since y=vx, we end up with \dfrac{\,dy}{\,dx}=v+x\dfrac{\,dv}{\,dx}

    Thus, the ODE becomes v+x\dfrac{\,dv}{\,dx}=\dfrac{v^3+1}{v^2}. This is now a separable equation.

    Can you proceed?

    EDIT: Man, am I slow!!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Nov 2009
    Posts
    69
    You are integrating with respect to v when you should be integrating with respect to x. You need to make the substitution \frac{dy}{dx} = v + x \frac{dv}{dx} and solve the resulting equation for v using separation of variables. Once you have v you can get y since y = vx.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member
    Joined
    Apr 2010
    Posts
    487
    Thanks guys!

    With the 3C in the solution, is it fair to just write 'C' since it's still an unknown constant?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,404
    Thanks
    1293
    Quote Originally Posted by Glitch View Post
    Thanks guys!

    With the 3C in the solution, is it fair to just write 'C' since it's still an unknown constant?
    I'd use a different letter just because \displaystyle 3C \neq C, but yes, it's still just an arbitrary constant.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Differential Equation Substitution
    Posted in the Differential Equations Forum
    Replies: 10
    Last Post: January 28th 2011, 09:39 AM
  2. differential equation with substitution!!!
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: February 4th 2009, 03:52 PM
  3. Differential Equation : y/x substitution
    Posted in the Differential Equations Forum
    Replies: 10
    Last Post: February 1st 2009, 08:54 AM
  4. Differential equation substitution
    Posted in the Differential Equations Forum
    Replies: 5
    Last Post: May 25th 2008, 10:50 AM
  5. substitution for differential equation
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 13th 2008, 05:29 PM

Search Tags


/mathhelpforum @mathhelpforum