Thread: Linear ODE / Bernoulli Equation

1. 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.

2. Re: Linear ODE / Bernoulli Equation

Originally Posted by cheme
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 \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 $\displaystyle \displaystyle v = \frac{y}{x} \implies y = vx \implies \frac{dy}{dx} = v + x\,\frac{dv}{dx}$ and the DE becomes

\displaystyle \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.

3. Re: Linear ODE / Bernoulli Equation

To get to this step,

\displaystyle \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.

4. Re: Linear ODE / Bernoulli Equation

Originally Posted by cheme
To get to this step,

\displaystyle \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 $\displaystyle \displaystyle 1$, in this case, $\displaystyle \displaystyle \frac{\frac{1}{x}}{\frac{1}{x}}$.

5. Re: Linear ODE / Bernoulli Equation

Originally Posted by Prove It
I multiplied by a cleverly disguised $\displaystyle \displaystyle 1$, in this case, $\displaystyle \displaystyle \frac{\frac{1}{x}}{\frac{1}{x}}$.
I don't know how you were able to see that, but good job. Thank you.

6. Re: Linear ODE / Bernoulli Equation

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