Question

bernoullis equation

Answer 1

\[6y^2dx−x(2x^3+y)dy=0\]

Answer 2

find the solution

Answer 3

the general
solution

Answer 4

http://tutorial.math.lamar.edu/Classes/DE/Bernoulli.aspx I know how to do it,but i dont get the right answer thats why i need help

Answer 5

i don't think that is a Bernoulli equation

Answer 6

i think you're right, sir. not Bernoulli

Answer 7

Solving for \(\dfrac{\mathrm dx}{\mathrm dy}\), we can see that the ODE is indeed a Bernoulli equation.
\[\begin{align*}6y^2\,\mathrm dx-x(2x^3+y)\,\mathrm dy=0\implies\quad \quad \quad\frac{\mathrm dx}{\mathrm dy}-\frac{1}{6y}x&=\frac{1}{3y^2}x^4\\[1ex]
x^{-4}\frac{\mathrm dx}{\mathrm dy}-\frac{1}{6y}x^{-3}&=\frac{1}{3y^2}
\end{align*}\]Set \(u=x^{-3}\), so \(-\dfrac{1}{4}\dfrac{\mathrm du}{\mathrm dy}=x^{-4}\dfrac{\mathrm dx}{\mathrm dy}\). The ODE becomes
\[\begin{align*}
-\frac{1}{4}\frac{\mathrm du}{\mathrm dy}-\frac{1}{6y}u&=\frac{1}{3y^2}\\[1ex]
\frac{\mathrm du}{\mathrm dy}+\frac{2}{3}u&=-\frac{4}{3y^2}
\end{align*}\]which is linear in \(u\).

Answer 8

you star!!