Question

3. Consider the system of equations: x' = ( 1 3 | -1 6 -2 * (a) [4 pts) Find the general solution. (b) [4 pts) Find the critical points or equilibrium solutions. Plot a few representive trajectories of the system in the phase plane. Indicate the direction of each trajectory using arrows.

Answer 1

3) The given differential system is given by X'= (30) where x = ( 2 ) and A=(3, 62) the system reduces to 4 = 3x + 6*2 = f (24,4) kg = - 24 - 2%, Egexo,X9). Making f (x, 22) = 0 and g(x, y) = 0. We get the fixed point Here 24 = %=0. So, (0,0) is the only fixed point. .. Now, I = I fx fxz 1-1361 la grigl 17 Now, at (0,0). I is - 13 6 = AB. ... Now, IB-2I=0. 17 -2 13-7 6 1=0 (2+2)(x-3) +6 = 0. 1-1 -2-41 = 29 2.(-0) .: 7=0, 1. clearly (oo) is unstable Node. Let x, be the eigen veetor corresponding a=1. We get- Ax, = x, (A-IX, = 0 [261x1 =10 24 +62 = 0.} = -24 -3x₂ = 0 J x+3x = 0 24 =-3X2. . If x = (-k). then ... If x2 = (-k) . then, x = 3k. . Now, X, - ( 24 ) = (3x) = K (3) where ke iR-{o} Let X be the eigen vector corresponding a=0 we get AX2 - 0 X2 ]] x + 2X=0 . 34 +6X = 0 1 -24 - 2x2 = 0 x=-242. If x=(-k). a then, x4=2k.

then X2 = ( 2 ) = (2x) = k (²) where KE IR-foy therefore the general Solution is a x= a. (2 et + Cz. (4) . eot where ci ca are barbitary - a (et + C. (2) constant Now, the phase potrait is given by