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

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The discussion centers on plotting isoclines for the first-order ordinary differential equation (ODE) y' = 2 * y^3 * x, specifically for slopes m = +1 and m = -1. The isoclines are derived from the equation m = 2 * y^3 * x, leading to the expressions y = ±(m/(2*x))^(1/3). For m = 1, the isocline simplifies to y = ±(1/(2*x))^(1/3), and for m = -1, it simplifies to y = ±(-1/(2*x))^(1/3). A key correction noted is that the ± symbol is unnecessary when taking the cube root.

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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
 
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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 [itex]\pm[/itex] because you're taking the cube root not the square root.
 

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