Answer:
Step-by-step explanation:
we have the following differential equations
[tex]\frac{dx}{dt}=-2y\\\frac{dy}{dt}=-2x\\[/tex]
by differentiating the second equation we have
[tex]\frac{d}{dt}(\frac{dy}{dt})=-2\frac{dx}{dt}\\\frac{d^{2}y}{dt^{2}}=-2\frac{dx}{dt}\\\frac{dx}{dt}=\frac{-1}{2}\frac{d^{2}y}{dt^{2}}[/tex]
and we replace dx/dt in the first equation
[tex]\frac{-1}{2}\frac{d^{2}y}{dt^{2}}=-2y\\\frac{d^{2}y}{dt^{2}}-4y=0[/tex]
and by using the characteristic polynomial
[tex]m^{2}+4=0\\m=\±2i[/tex]
the solution is
[tex]y(t)=Acos(2t)+Bsin(2t)[/tex]
and to compute x(t) we have
[tex]\frac{dx}{dt}=-2Acos(2t)-2Bsin(2t)\\\\\int dx = \int[-2Acos(2t)-2Bsin(2t)]dt\\\\x(t)=-Asin(2t)+Bcos(2t)[/tex]
and if we use x(0)=4 and y(0)=3, we can calculate the constants A and B
[tex]x(0)=B=4\\y(0)=A=3[/tex]
I hope this is useful for you
regards
