Question In each of Problems 10 through 12, solve the given initial value problem. Describe the behavior of the solution as t → 0. 10. x = (3 - 7)*, x0) = (-3) 11. x = ( 1 ) + x(0) = (2)
need help on both 10 and 11
(10) Given x^(')(t)=([5,-1],[3,1])x,x(0)=([2],[-1])solve homogeneous system: A=([5,-1],[3,1])To find the eigen values solve det(A-lambda I)=0|[5-lambda,-1],[3,1-lambda]|=0=>(5-lambda)(1-lambda)+3=0=>lambda^(2)-6lambda+8=0=>(lambda-4)(lambda-2)=0Therefore the eigen values are lambda=2,4.To find the eigenvectors solve (A-lambda I)X=0: For lambda=2 :([5-2,-1],[3,1-2])X=0=>([3,-1],[3,-1])X=0=>([1,-1//3],[0,0])X=0The corresponding system of equations is: x-(1)/(3)y=0=>x=(1)/(3)y.Here, the free variable is y=s, so, x=(1)/(3)s. Then ([x],[y])=([s//3],[s])=s([1//3],[1]).The corresponding eigenvector is ([1//3],[1]). Then one solution will be x_(1)(t)=([1//3],[1])e^(2t).For lambda=4:([5-4,-1],[3,1-4])X=0=>([1,-1],[3,-3])X=0=>([1,-1],[0,0])X=0The corresponding system of equations is: x-y=0=>x=y.Here, the free variable is y=s, so, x=s, Then ([x],[y])=([s],[s])=s([1],[1]).The corresponding eigenvector is ([1],[1]). Then one solution will be x_(2)(t)=([1],[1])e^(4t).Then the general solution of homogeneous is x_(c)(t)=c_(1)([1//3],[1])e^(2t)+c_(2)([1],[1])e^(4t).To find c_(1),c_(2) use x(0)=2,y(0)=-1 :x(0)=c_(1)([1//3],[1])e^(0)+c_(2)([1],[1])e^(0)=>([2],[-1])=([(1)/(3)c_(1)+c_(2)],[c_(1)+c_(2)])=>{[(1)/(3)c_(1)+c_(2)=2],[c_(1)+c_(2)=-1]:}Solving we get, c_(1)=-(9)/(2),c_(2)=(7)/(2).So, x(t)=-(9)/(2)([1//3],[1])e^(2t)+(7)/(2)([1],[1])e^(4t)(11) ... See the full answer