Question

A solution curve has been superimposed on the slope field shown below:

y’ = tan(x); y(0) = 0
y’ = cot(x); y(pi/4) = 1
y’ = tan(x); y(0) = 0
y' = 1/(1+x^2); y(pi/4) = 1
y’ = 1 + y^2; y(0) = 0

Answer 1

@zepdrix

Answer 2

we could work backwards, the graph shows a curve
y = tan(x)

y' = sec^2 (x)

we know that
sec^2 = 1 + tan^2

--> y' = 1 + y^2