Resolved: "Solution of simultaneous ODE :x′=(x2+y2)y and y′=−(x2+y2)x"

High School
Calculus and Analysis
Calculus 2
Differential equations

Dania Robbins

Answered question

2022-04-23

Solution of simultaneous ODE :

x^{'} = (x^{2} + y^{2}) y

and

y^{'} = - (x^{2} + y^{2}) x

Answer & Explanation

tswe0uk

Beginner2022-04-24Added 19 answers

Step 1
You can assemble the two equations into a complex-scalar equation as
$\dot{z} = - i {| z |}^{2} z, z = x + i y .$
As observed, the radius has then the equation
$r \dot{r} = \frac{12}{\frac{d}{d t}} {| z |}^{2} = R e (\overset{―}{z} \dot{z}) = R e (- i) {| z |}^{4} = 0,$
giving $r = c$ constant. Then the original equation has a simple linear form
$\dot{z} = λ z, λ = - i c^{2}$
$\Rightarrow z (t) = z_{0} e^{- i c^{2} t}, | z_{0} | = c or z_{0} = c e^{i ϕ}$
$\Rightarrow x (t) + i y (t) = c, (\cos (c^{2} t - ϕ) - i \sin (c^{2} t - ϕ))$

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