(a) Find the general solution for x' = [ 1 -2 3 -4 ] x. (b)...

Transcript

00:01 Hello everyone here the given problem is in this problem x -dash this is equal to matrix 1 minus 2 3 minus 4 into x so this is our first equation now x 1 dash x2 this is equal to matrix 1 minus 2 3 minus 4 into x is x1 and x2 which implies x1 dash x2 this is equal to x1 minus 2 x2 this is equal to x1 minus 4 x2 okay we multiply these two matrix then x1 this is equal to x1 minus 2 x2 this is our third equation and x2 this is equal to 3 of x1 minus 4 of x2 this is our fourth equation let d by d t this is equal to d from third and fourth from third and fourth equation d x1 divided by d t this is equal to x1 minus 2 x2 which implies d minus 1 of x1 plus 2 x2 this is equal to 0 so d x2 divided by dd this is equal to 3x1 minus 4 x2 which implies d plus 4 x2 minus 3x1 this is equal to 0 so therefore, d1 minus 1 of x1 plus 2x2, this is equal to 0.

02:34 Now, this is equation 5th, okay? d plus 4 x2 minus 3x1, this is equal to 0, which implies minus 3x1 plus 1 plus d plus d plus 4.

02:55 X2 this is equal to 0 this is our sixth equation from fifth and sixth three into d minus 1 x1 plus 6x2 this is equal to 0 minus 3 d minus 1 x1 plus d minus 1 x1 into x2 this is equal to 0 let this is our 7 equation this equation okay and this is our eight equation so adding equation adding equation seven and eight we get we have d plus four d minus one plus six into x2 this is equal to zero and d square minus d plus four d minus four plus six into x2 this is equal to 0 now d square plus 3d plus 2 into x2 this is equal to 0 so we can write this m square plus 3m plus 2 this is equal to 0 so m square plus 2m plus m plus 2 we can write this we can split this middle time now here we take open am so the here is m plus 2 plus 1 here we take common 1 then the the left term is m plus 2 so this is equal to 0 so from here sorry m plus 1 and m plus 2 this is equal to 0 we can say that m plus 1 this is also equal to 0 so from here we can fight the value of m is minus 1 and m plus 2 this is equal to 0 so from here we can write the value of m is equal to m is equal to minus 2 so therefore x2 this is equal to c1 e 2 power minus t plus c2 e raise 2 power minus 2 t let this is our ninth equation sorry this is our 9 equation now using 9 in 6 we have minus 3x1 plus d plus 4 x2 this is equal to 0 so the value of 3x1 this is equal to d plus 4 x2 so this is equal to minus c1 he raised 2 power minus t plus c2 minus 2 into minus 2 and e raise 2 power minus 2 t plus 4 c1 e raised 2 power minus t plus plus 4 c2 e raise 2 power minus t so 3 x1 this is equal to 3 c1 e raise 2 power minus t plus 2 c2 e raise 2 to 2 t x1 is equal to the value of x1 this is equal to c1 e raise 2 power minus t plus 2 by 3 c2 e raised 2 power minus 2 t let this is our 10 equation therefore x1 x2 this is equal to c1 e raise 2 power minus t plus 2 by 3 c2 e raise 2 2 2 t and here is c1 e raise 2 power minus t plus c2 e raise 2 power minus t plus c2 e x2 this x1 x2 this is equal to c1 e raise 2 minus t and e raised 2 power minus t okay and plus c 2 common then here is 2 by 3 a raise 2 power minus 2 t and here is a raise 2 power minus 2 t okay now x1 x2 this is equal to c1 here is 1 1 1 e -rease 2 -1 -t2 -mine -t take common plus c2 here is 2 -5 -3 and here is 1 so here we take common e -re -to -mine minus 2 t then the general solution is general solution is x of t this is equal to c1 1 1 1 e -re 2 power minus t plus c 2 2 by 3 1 e raised 2 power minus to t okay therefore the linearly independent solution is linearly independent solution is 3 so x1 this is equal to alpha i raise to power minus t and x2 this is equal to beta e raise 2 power minus 2 t okay where where alpha is equal to 1 1 beta is equal to 2 by 3 1 let x 1 x2 this is equal to determinant of x1 x2 dash x2 dash x2 okay so this is equal to minus alpha raised to power minus t here this is alpha erase to power minus t and here is minus two beta erase to power minus two t and here is beta e raise to power minus two t okay now this is equal to minus alpha beta e raise to power minus 3 t plus 2 alpha beta e raise to power minus 3 t plus 2 alpha beta e raise to power minus 3 t this is equal to to minus alpha beta a raised to power minus 3 t this is not equal to 0 and e raised to power minus 3 t this is not equal to 0 and alpha this is not equal to 0 and beta this is also not equal to 0 here x1 and x2 x1 and x2 are linearly independent are linearly nearly independent...

Question

(a) Find the general solution for x' = [ 1 -2 3 -4 ] x. (b) Show whether the Wronskian of the two solutions you found is zero or nonzero. Do they form a fundamental set of solutions? (c) Explain the behavior of the solutions as t ? ? if the initial condition is given by x(0) = [ 1 2 ].

Question

(a) Find the general solution for x' = [ 1 -2 3 -4 ] x. (b) Show whether the Wronskian of the two solutions you found is zero or nonzero. Do they form a fundamental set of solutions? (c) Explain the behavior of the solutions as t ? ? if the initial condition is given by x(0) = [ 1 2 ].

Please give Ace some feedback