Plotting isoclines corresponding to slopes m = +1 and m -1?

[sinbad30]

Given the first order ODE y' = 2 * y^3 * x, plot the isoclines corresponding to slopes m = +1 and m = -1.

My answer:

The isoclines are given by m= 2 * y^3 * x or y=(m/(2*x))^1/3

The slope elements on the isocline y=+- (m/(2*x))^1/3 all have gradient m

For m = 1, the isocline is y = +- (1/(2*x))^1/3
For m = -1, the isocline is y = +- (-1/(2*x))^1/3

and hence l simply plotted the above two equations

Can anyone confirm whether or not l am correct

Thanks

 

[Natr0n]

Seems correct except in

For m = 1, the isocline is y = +- (1/(2*x))^1/3
For m = -1, the isocline is y = +- (-1/(2*x))^1/3

You don't need \pm because you're taking the cube root not the square root.