Carole Yarbrough

2021-12-20

Solve the following differential equations. Use the method of Bernoulli’s Equation. $x\left(2{x}^{3}+y\right)dy-6{y}^{2}dx=0$

Bob Huerta

Step 1
General Bernoullis

Mason Hall

Solve $-x\frac{dy\left(x\right)}{dx}\left(2{x}^{3}+y\left(x\right)\right)+6y{\left(x\right)}^{2}=0$:
Solve for $\frac{dy\left(x\right)}{dx}$:
$\frac{dy\left(x\right)}{dx}=\frac{6y{\left(x\right)}^{2}}{x\left(2{x}^{3}+y\left(x\right)\right)}$
Write the differential equation in terms of x. Since $\frac{dy}{dx}\frac{dx}{dy}=1,\frac{dy\left(x\right)}{dx}=\frac{1}{\frac{dx\left(y\right)}{dy}}$:
$\frac{1}{\frac{dx\left(y\right)}{dy}}=\frac{6{y}^{2}}{\left(y+2x{\left(y\right)}^{3}\right)x\left(y\right)}$
Raise both sides to the power -1 and expand:
$\frac{dx\left(y\right)}{dy}=x\frac{{\left(y\right)}^{4}}{3{y}^{2}}+x\frac{y}{6y}$
Subtract $x\frac{y}{6y}$ from both sides:
$\frac{dx\left(y\right)}{dy}-x\frac{y}{6y}=x\frac{{\left(y\right)}^{4}}{3{y}^{2}}$
Divide both sides by $-\frac{1}{3}x{\left(y\right)}^{4}$:
$-\frac{3\frac{dx\left(y\right)}{dy}}{x}{\left(y\right)}^{4}+\frac{1}{2yx{\left(y\right)}^{3}}=-\frac{1}{{y}^{2}}$
Let $v\left(y\right)=\frac{1}{x}{\left(y\right)}^{3}$, which gives $\frac{dv\left(y\right)}{dy}=-\frac{3\frac{dx\left(y\right)}{dy}}{x}{\left(y\right)}^{4}$

RizerMix

The equation $6{y}^{2}dx-x\left(2{x}^{3}+y\right)dy=0$ , for y0 can be written as the following Bernulli equation in the unknown x(y)
$dx/dy-x/6y={x}^{4}/3{y}^{2}$. To reduce the equation to a linear one take $x=V\left(y{\right)}^{-1/3}$. Obtain ${V}^{\prime }+V/2y=-1/{y}^{2}$. The integrating factor is ${y}^{1/2}$ and the solution is .
Obtain the solution as ${x}^{3}=y/\left(2+C{y}^{1/2}\right)$ or $\left(2{x}^{3}-y{\right)}^{2}=Ky{x}^{6},K={C}^{2}$

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