Question

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I AM CURRENTLY IN DIFFERENTIAL EQUATIONS

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Consider the linear system x' = Px where P is a 2 x 2 constant matrix. Assume P has only one eigenpair (-2, Then the general solution of this system might have the form: (-2. (

Answer 1

x= px, one eigen -26 -2t -26 -26 -26 the Second Linearly independant solution is in and given p has only pair 62, [o]) ine, 12-2, VE [b] the form of viteette X(t)= (vit +Valē ut xe) > x'ct) = 6 2014–2 Utvidelse and Ax = (tay+Al) (AV2 +2VZ) = 4 (A +2I) I 2 = 0 this equation will give V₂ (i is known find A =P (I intentionly taleen = (uit+V2 ]2 Settuale tue = (-2V₂ tv, -zuit) -2 22t e 2 by compare ☺ e AM -2 Vi and AV2 C-2 V2+VI Now the problem is but it is A sponly the eigen values of Ap. are -21-2 lit pa la d] atd=-2-2=-4 (trace a sum of and (A +2I) v = eigenvaluey) and ad-bC=2.02=4, ((P1= product of eighs parte din 603-609 valery) glaarz and then 3-4 at2 =0 ) c=0 and adobe ao b is any valie. 3. p= to be

(P +2I) v = ". L :][] = [.] os ba2=0 it bto, n2=0 &x=kto it is already V, bzo © :]() = 2] m=k, 192382 02= [0]k + [i]ke for kizo for kizo, take U2= take Uza 0 kzal the another solution is uiteet ovzelt (uit+uz) ezt ([b]++[*]]ēl [tjelt. the Solution is xa)= ei [66]+ ſtēt : = - [ter ezt ezt Ju