Question

Q2: Find the general solution to T' = AT where A is given by 1-3 -2] A = 2 1]' and then sketch a variety of solution curves.

Answer 1

Here on ZAR where A = [-3 -2] we can rewrite the system as and de 1 2 11=-3+2 de 2-3x, -20 e for equilibrium point is given by getting dress to and free to 20 22x1+12 . . -311- 24.220 - and 2011 f1220 then the coofficient determinant therefore, (016) is the only of the system is 1-3 3 = -3+2 equilibrium point. 2-110 let, xi=dent and xe = Bett be a solution of the system, then the characteristic equation is det (-I) - |-3-1 -21. | 21- 30 = (-3-1)(1-) - 64) 20 >> +3) 6-1) +4 20 7 ²+ 37-7-3 +420 => 3?+27 +120 =9+)30 therefore the eigen values ane )=-1,-1 21=-1 and 722-1 Now, 2 (3-1) + (-2) B=0 -- 0 20 +(1-0) B20 -- if the solution na dent and Rza Bedt wsing equation putting Di=-1 we get (-3-di) d+E2) B:2 » +3+) a +(2) B 2o ~ -20 +62) B is the respective eigen veetoreign ñ kl ) » daß, if a=k , B=k

Next step is to find 7 from ① and w as 22-1 we need to solve 5) -241-28291 » 20,+ 2.822-1 » Batdzak = B22-d2t when drak, B2 = -K-1 .. 2 d220 the general solution for the systen is cet (1) +8{te*(1)+c+()) which is the and required the Phase general Potrait solution , is since *1=2224 <o

the solution curve is singular node or stable singular node.