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Math Help - Linear ODE / Bernoulli Equation

  1. #1
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    Linear ODE / Bernoulli Equation

    I am having trouble solving the following differential equation:

    4xy + 6y^2 + (2x^2 + 6xy)y' = 0

    I have tried to bring it into the form y' + p(x)y = r(x) but cannot, thereby thinking it is a Bernoulli Equation. When I try to put it into Bernoulli form and get the y' by itself, I am left with the (2x^2 + 6xy) in the denominator of the other terms, making it impossible to solve.
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  2. #2
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    Re: Linear ODE / Bernoulli Equation

    Quote Originally Posted by cheme View Post
    I am having trouble solving the following differential equation:

    4xy + 6y^2 + (2x^2 + 6xy)y' = 0

    I have tried to bring it into the form y' + p(x)y = r(x) but cannot, thereby thinking it is a Bernoulli Equation. When I try to put it into Bernoulli form and get the y' by itself, I am left with the (2x^2 + 6xy) in the denominator of the other terms, making it impossible to solve.
    \displaystyle \begin{align*} 4xy + 6y^2 + (2x^2 + 6xy)\,\frac{dy}{dx} &= 0 \\ (2x^2 + 6xy)\,\frac{dy}{dx} &= -4xy - 6y^2 \\ \frac{dy}{dx} &= \frac{-4xy - 6y^2}{2x^2 + 6xy} \\ \frac{dy}{dx} &= \frac{-2y(2x + 3y)}{2x(x + 3y)} \\ \frac{dy}{dx} &= -\left(\frac{y}{x}\right)\left(\frac{2x + 3y}{x + 3y}\right) \\ \frac{dy}{dx} &= -\left(\frac{y}{x}\right)\left[\frac{2 + 3\left(\frac{y}{x}\right)}{1 + 3\left(\frac{y}{x}\right)}\right]\end{align*}

    Now make the substitution \displaystyle v = \frac{y}{x} \implies y = vx \implies \frac{dy}{dx} = v + x\,\frac{dv}{dx} and the DE becomes

    \displaystyle \begin{align*} v + x\,\frac{dv}{dx} &= -v\left(\frac{2 + 3v}{1 + 3v}\right) \\ x\,\frac{dv}{dx} &= -v\left(\frac{2 + 3v}{1 + 3v}\right) - v \end{align*}

    This is now a separable DE.
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    Re: Linear ODE / Bernoulli Equation

    To get to this step,

    \displaystyle \begin{align*}\\ \frac{dy}{dx} &= -\left(\frac{y}{x}\right)\left[\frac{2 + 3\left(\frac{y}{x}\right)}{1 + 3\left(\frac{y}{x}\right)}\right]\end{align*}

    did you divide each term in the parenthesis by x? Algebraically, I am not sure how you got to that step.
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    Re: Linear ODE / Bernoulli Equation

    Quote Originally Posted by cheme View Post
    To get to this step,

    \displaystyle \begin{align*}\\ \frac{dy}{dx} &= -\left(\frac{y}{x}\right)\left[\frac{2 + 3\left(\frac{y}{x}\right)}{1 + 3\left(\frac{y}{x}\right)}\right]\end{align*}

    did you divide each term in the parenthesis by x? Algebraically, I am not sure how you got to that step.
    I multiplied by a cleverly disguised \displaystyle 1, in this case, \displaystyle \frac{\frac{1}{x}}{\frac{1}{x}}.
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    Re: Linear ODE / Bernoulli Equation

    Quote Originally Posted by Prove It View Post
    I multiplied by a cleverly disguised \displaystyle 1, in this case, \displaystyle \frac{\frac{1}{x}}{\frac{1}{x}}.
    I don't know how you were able to see that, but good job. Thank you.
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  6. #6
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    Re: Linear ODE / Bernoulli Equation

    Quote Originally Posted by cheme View Post
    I don't know how you were able to see that, but good job. Thank you.
    Since I was making the substitution \displaystyle v = \frac{y}{x}, I knew I had to get a function of \displaystyle \frac{y}{x}.
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