Solve the given nonlinear plane autonomous system by changing to polar coordinates.(r(t), 0(t))x' = y -y': = -X-X(0) = (1, 0)(r(t), 0(t))4 13e8t=X3e St5X(0) = (4,0)+ y²yx² + y²(16 - x² - y²)+12Describe the geometric behavior of the solution that satisfies the given initial condition.The solution approaches the origin on the ray 8 = 0 as t increases.The solution spirals toward the circle r = 4 as t increases.The solution traces the circle r = 4 in the clockwise direction as t increases.The solution spirals away from the origin with increasing magnitude as t increases.The solution spirals toward the origin as t increases.(16 - x² - y²),(solution of initial value problem)X)(solution of initial value problem)Describe the geometric behavior of the solution that satisfies the given initial condition.The solution approaches the origin on the ray 0 = 0 as t increases.The solution spirals toward the circle r = 4 as t increases.The solution traces the circle r = 4 in the clockwise direction as t increases.The solution spirals away from the origin with increasing magnitude as t increases.The solution spirals toward the origin as t increases.

Given: Nonlinear plane autonomous system of equations as- x'=y-x(x2+y2)16-x2-y2y'=-x-y(x2+y2)16-x2-y2, X0=1, 0To find: Polar coordinates rt,θtWe have given the system of equations as-x'=y-x(x2+y2)16-x2-y2y'=-x-y(x2+y2)16-x2-y2, X0=1, 0In polar coordinates, we know that-r2=x2+y2-1 tan θ=yx -2On differentiating 1 w.r.t "t", dr2dt=ddtx2+y2⇒2rr'=2xx'+2yy'⇒r'=xx'+yy'r =xy-xx2+y216-x2-y2+y-x-yx2+y216-x2-y2r =xy-x2x2+y216-x2-y2-xy-y2x2+y216-x2-y2r =-x2x2+y216-x2-y2-y2x2+y216-x2-y2r =-16-x2-y2x2x2+y2+y2x2+y2r =-16-x2+y2x2+y2x2+y2r =-16-r2r2r2r =-16-r2 =r2-16 ⇒r'=r2-16On differentiating 2 w.r.t "t", dtanθdt=ddtyx⇒sec2θ dθdt=xy'-yx'x2⇒1+tan2θ dθdt=xy'-yx'x2⇒1+y2x2 dθdt=xy'-yx'x2⇒x2+y2x2 dθdt=xy'-yx'x2⇒ dθdt=x-x+y(x2+y2-yy+x(x2+y2)x2+y2⇒ dθdt=-x2-y2x2+y2⇒ dθdt=-x2+y2x2+y2⇒ dθdt=-1⇒ θ'=-1Now,r'=r2-16⇒drdt=r2-16At t=0, x=1, y=0 givenOn integrating w.r.t. "t",we get- ∫drr2-16=∫dt⇒∫drr+4r-4=∫dt⇒∫18drr-4-drr+4=∫dt⇒∫drr-4-drr+4=8∫dt⇒logr-4r+4=8t+log c⇒r-4r+4=e8t+log c⇒r-4r+4=Ce8tUsing initial condition,X0=1,0;⇒1-41+4=Ce0⇒-35=C⇒C=-35∴r-4r+4=-35e8t⇒r-4+r+4r+4-r-4=-35e8t+11+35e8t⇒2r8=1-35e8t1+35e8t⇒r=41-35e8t1+35e8tAlso,dθdt=-1At t=0, θ=0;On integrating w.r.t. "t",we get-∫0θdθ=-∫0tdt⇒θ-0=-t-0⇒θ=-t∴rt,θt=41-35e8t1+35e8t12,-tThe clockwise flow on the circle of radius r=4 describes a trajectory that goes inwards as a spiral for r'<0 and outwards for r'>0 as t increases.