Question

Solve the following differential equation

Answer 1

\[\frac{ dy }{ dx }=\frac{ x^{2}-y^{5} }{ 2xy+x^{3}}\]

Answer 2

@phi

Answer 3

I dont think this can be solved explicitly.
Here is a reference to how to estimate the graph of the solution:

http://tutorial.math.lamar.edu/Classes/DE/EulersMethod.aspx

Answer 4

That's applicable when initial condition is given. But in the given problem no such initial condition is provided.

Answer 5

Possible approach you can consider:

Rewrite the ODE.\[\frac{\mathrm dy}{\mathrm dx}=\frac{ x^{2}-y^{5} }{ 2xy+x^{3}}\iff(y^5-x^2)\,\mathrm dx+(2xy+x^3)\,\mathrm dy=0\]We can try looking for a solution of the form \(\Psi(x,y)=C\), which by the chain rule gives
\[\mathrm d\Psi=\color{red}{\frac{\partial \Psi}{\partial x}}\,\mathrm dx+\color{blue}{\frac{\partial \Psi}{\partial y}}\,\mathrm dy=\color{red}{M(x,y)}\,\mathrm dx+\color{blue}{N(x,y)}\,\mathrm dy=0\]If the mixed partial derivatives are equal, the ODE is called exact.

That's not case here, as you have
\[\begin{align*}
\frac{\partial^2\Psi}{\partial y\,\partial x}&=\frac{\partial}{\partial y}\left[y^5-x^2\right]=5y^4\\[1ex]
\frac{\partial^2\Psi}{\partial x\,\partial y}&=\frac{\partial}{\partial x}\left[2xy+x^3\right]=2y+3x^2
\end{align*}\]But it may be possible to find what's called an integrating factor which may be a function of \(x\), \(y\), or both.

In the first case, we're looking for a function \(\mu(x)\) that makes the ODE
\[\mu(x)M(x,y)\,\mathrm dx+\mu(x)N(x,y)\,\mathrm dy=0\]exact, which would require
\[\begin{align*}
\frac{\partial}{\partial y}\left[\mu(x)M(x,y)\right]&=\frac{\partial}{\partial x}\left[\mu(x)N(x,y)\right]\\[1ex]
\mu M_y&=\frac{\mathrm d\mu}{\mathrm dx}N+\mu N_x\\[1ex]
N\frac{\mathrm d\mu}{\mathrm dx}&=\mu(M_y-N_x)\\[1ex]
\frac{\mathrm d\mu}{\mu}&=\frac{M_y-N_x}{N}\,\mathrm dx\\[1ex]
\implies \mu(x)&=\exp\left(\int\frac{M_y-N_x}{N}\,\mathrm dx\right)
\end{align*}\]Similarly, if \(\mu\) is a function of \(y\), you can derive an integrating factor of the form
\[\mu(y)=\exp\left(\int\frac{N_x-M_y}{M}\,\mathrm dy\right)\]Either integrating factor would have to be a function of their respective variable, but that's not the case here, so you can try finding an integrating factor in terms of both variables. For that, you need to find \(\mu(x,y)\) such that
\[\begin{align*}
\frac{\partial}{\partial y}\left[\mu(x,y)M(x,y)\right]&=\frac{\partial}{\partial x}\left[\mu(x,y)M(x,y)\right]\\[1ex]
\mu_yM+\mu M_y&=\mu_xN+\mu N_x\\[1ex]
\mu_xN-\mu_yM&=\mu(M_y-N_x)\\[1ex]
\frac{\mu_x}{\mu}N-\frac{\mu_y}{\mu}M&=M_y-N_x\end{align*}\]Probably the easiest solution we can hope for is a separable one, that is one of the form \(\mu(x,y)=f(x)g(y)\):
\[\begin{align*}
\frac{f'(x)g(y)}{f(x)g(y)}N-\frac{f(x)g'(y)}{f(x)g(y)}M&=M_y-N_x\\[1ex]
\frac{f'(x)}{f(x)}N-\frac{g'(y)}{g(y)}M&=M_y-N_x\\[1ex]
F(x)N-G(y)M&=M_y-N_x
\end{align*}\](because both of \(\dfrac{\mu_x}{\mu}\) and \(\dfrac{\mu_y}{\mu}\) are functions of only \(x\) and \(y\), respectively).

From here, we want to find \(F(x)\) and \(G(y)\) such that
\[F(x)(2xy+x^3)+G(y)(y^5-x^2)=5y^4-2y-3x^2\]This in general isn't easy, however. One way you can proceed is to try some power/polynomial guesses for \(F\) and \(G\). I tried \(F=ax^\alpha\) and \(G=by^\beta\) but that didn't seem to work out. You may have better luck.