Can anyone please help me how to find the solution for: $$ \frac{\mathrm dy}{\mathrm dx}=(x-5y)^{\frac{1}{3}}+ \frac{1}{5}. $$ I found the singular curve for it, which is $y=\dfrac{x}{5}$ (please correct me if I'm wrong), but I can not find the family of solution for it. I tried every method that I know. Please help me.
Thank you in advance for your assistance.
On
$$\frac{\mathrm dy}{\mathrm dx}=(x-5y)^{\frac{1}{3}}+ \frac{1}{5}.$$ $$5y'=5(x-5y)^{\frac{1}{3}}+1$$ $$-(1-5y')=5(x-5y)^{\frac{1}{3}}$$ You observe that the derivative of $x-5y$ is just $1-5y'$
$$\int \frac {d(x-5y)}{(x-5y)^{\frac{1}{3}}}=-5\int dx$$
It's easy to solve now $$ \frac 32{(x-5y)^{\frac{2}{3}}}=-5x+K$$
Note that if \begin{align} u:= x-5y \end{align} then \begin{align} \frac{du}{dx} = 1- 5\frac{dy}{dx}. \end{align} Using this fact, we can rewrite the differential equation as \begin{align} \frac{1}{5}\frac{du}{dx} = \frac{1}{5}\left(1-5\frac{dy}{dx} \right) =\frac{1}{5}-\frac{dy}{dx} = -(x-5y)^{1/3} =-u^{1/3}. \end{align} Hence we have a separable equation.