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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.

Respuesta :

Answer:[tex]2y^2 = x^2+4[/tex]

Step-by-step explanation:

given is a differntial equation in parametric form as

[tex]x'=-4y\\y'=-2x[/tex]

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

[tex]\frac{dy}{dx} =\frac{-2x}{-4y} \\4ydy =2x dx\\2y^2 = x^2+C[/tex]

(By separating the variables and integrating)

C is arbitrary constant

When x=-2, y =-2

Substitute to get

[tex]8=4+C\\C=4[/tex]

So solutoin is

[tex]2y^2 = x^2+4[/tex]

is the solution

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