Question #349893

dx dt=2x-4y

dx dt=2x+4y

Expert's answer

"Dx-2x+4y=0""Dy-2x-4y=0"

"(D-2)x+4y=0""-2x+(D-4)y=0"

Multiply first equation by "2" and second equarion by "D-2"

Add two equations

"+(-2(D-2)x+(D-2)(D-4)y)=0"

Simplify

"D^2-6D+16y=0"Auxiliary equation

"r=3\\pm i\\sqrt{7}"

"y(t)=c_1e^{3t}\\cos(\\sqrt{7}t)+c_2e^{3t}\\sin(\\sqrt{7}t)"

Then

"-\\sqrt{7}c_1e^{3t}\\sin(\\sqrt{7}t)+\\sqrt{7}c_2e^{3t}\\cos(\\sqrt{7}t)"

Substitute

"x=-2y+\\dfrac{1}{2}\\dfrac{dy}{dt}"

"+\\dfrac{3}{2}c_1e^{3t}\\cos(\\sqrt{7}t)+\\dfrac{3}{2}c_2e^{3t}\\sin(\\sqrt{7}t)"

"-\\dfrac{\\sqrt{7}}{2}c_1e^{3t}\\sin(\\sqrt{7}t)+\\dfrac{\\sqrt{7}}{2}c_2e^{3t}\\cos(\\sqrt{7}t)"

"+(-\\dfrac{1}{2}c_2-\\dfrac{\\sqrt{7}}{2}c_1)e^{3t}\\sin(\\sqrt{7}t)"

"y(t)=c_1e^{3t}\\cos(\\sqrt{7}t)+c_2e^{3t}\\sin(\\sqrt{7}t)"

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