How to graph directional field of nonlinear first order ODE and higher order DE's? For example, if we had this nonlinear first order ODE:
$$(y')^2+5yy'-10=0$$
How could we plot the directional field of it ?
And what about higher order DE, could we plot directional field to them ? I mean, if some ODE is really hard or even impossible to solve exactly, could we plot it as we do for first order?
Thanks.
 A: Plotting the directional field in general can be "approximated" for any arbitrarily complex first order ODE.
Consider the generic ODE
$$ F(x,y,y') = 0$$ 
To draw a direction field we need to calculate the value of $y'$ at each point $(x_p, y_p)$ in the plane. That amounts to solving
$$ F(x_p,y_p, y') = 0$$ 
For each pair of points, and using those slopes to draw the correct lines. That approach makes just as much sense here. For example at the point (1,3) we have 
$$ (y')^2 +5(3)y'-10 = 0 \rightarrow y' = \frac{-15 \pm \sqrt{265}}{2}$$
So there are "two" possible slopes in the slope field diagram of this function. (In fact there are two possible diagrams that are smoothly changing, depending on the sign of the radical you pick). 
Now for certain equations such as
$$ e^{e^{y'}} + (y')^3 - \sin(y+x^{y'}) = 0$$
It might not be algebraically feasible to find a closed form for $y'$ given an $x_p, y_p$. In which case we need to APPROXIMATE it, through some method of your choice (all of which are just fine tuned and intelligent versions of guess-and-check). 
Now the discussion of slope fields for higher order differential equations (meaning larger derivatives) doesn't really make sense unless you generalize the underlying object.
What does a slope field really show? Recall the solution to $F(x,y,y')$ will be of the form $y = G(x,C)$ where $C$ is some constant of choice (a degree of freedom). A slope field shows for each curve, generated by a choice of C, what the resulting graph "flows" and looks like.
For higher order differential equations $F(x,y,y',y'' ... y^{(n)}) = 0$ instead of a field we have a 2,3... n dimensional space of possible choices of constants. Since the solution will be of the form $y = G(x,C_1, C_2 ... C_n)$. Now here one can again make sense of drawing slopes in the conventional sense. 
A: Just solve $y'=y'(y)$ and you will get the various vector fields that are associated to the equation (there are at least two, the computations will tell you if there are more).
Of course this depends on solving a quadratic equation, in your example. It is unlikely to succeed in general equations, besides of course getting often nonunique vector fields.
The other aspect of the matter is that it is really not a good idea to proceed as you suggest. It is like expecting that there is a general well-organized procedure to obtain a bifurcation diagram. There is not. Similarly, particularly due to the nonuniqueness (see above) besides not being able (easily) to draw a phase portrait the truth is that the properties of the solutions may vary substantially.
