Question

Find the solution of the system x′=−4y,y′=−2x, where primes indicate derivatives with respect to t, that satisfies the initial condition x(0)=−2,y(0)=−2.

Answer 1

Answer:$2y^2 = x^2+4$

Step-by-step explanation:

given is a differntial equation in parametric form as

$x'=-4y\\y'=-2x$

We can find dy/dx by dividing y' by x'

$\frac{dy}{dx} =\frac{-2x}{-4y} \\4ydy =2x dx\\2y^2 = x^2+C$

(By separating the variables and integrating)

C is arbitrary constant

When x=-2, y =-2

Substitute to get

$8=4+C\\C=4$

So solutoin is

$2y^2 = x^2+4$

is the solution