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Consider a linear system dx→dt=Ax→of arbitrary size. Suppose x1(t)and x2(t)are a solution of the system. Is the sumx→(t)=x→1(t)+x→2(t) a solution as well? How do you know?

Short Answer

Expert verified

Yes,x→(t)=x→1(t)+x→2(t) is a solution of the linear system.

Step by step solution

01

Determine the equation of the solution.

Consider x→1tand x→2tis a solution of the linear system dx→dtwhere A is n×nmatrix.

Assume λ1,λ2,λ3,...,λnbe the Eigen values of the A matrix then there exist Eigen vectors v→1,v→2,v→3,..,v→nsuch that x→1(t)=c1eλ1,tv→1+c2eλ2,tv→2+...+cneλn,tv→nand x→2(t)=b1eλ1tv→1+b2eλ2tv→2+...+bneλntv→nwhere c1,c2,.......,cnand b1,b2,...,bnare constant.

If x(t) is the solution of the linear system dx→dt=Ax→then x→(t)=c1eλ1,tv→1+c2eλ2,tv→2+...+cneλn,tv→nwhereλ1,λ2,λ3,...,λnbe the Eigen values andv→1,v→2,v→3,...,v→nof n×nmatrix A.

02

Show that   is a solution of the system.

Adding the equations x→1(t)=c1eλ1,tv→1+c2eλ2,tv→2+...+cneλn,tv→nand x→2(t)=b1eλ1tv→1+b2eλ2tv→2+...+bneλntv→nas follows:

x→1(t)+x→2(t)=c1eλ1tv→1+c2eλ2tv→2+...+cneλntv→n+b1eλ1tv→1+b2eλ2tv→2+...+bneλntv→n

Simplify the equation

x→1(t)+x→2(t)=c1eλ1tv→1+c2eλ2tv→2+...+cneλntv→n+b1eλ1tv→1+b2eλ2tv→2+...+bneλntv→nas follows.

x→1(t)+x→2(t)=c1eλ1tv→1+c2eλ2tv→2+...+cneλntv→n+b1eλ1tv→1+b2eλ2tv→2+...+bneλntv→nx→1(t)+x→2(t)=c1eλ1tv→1+c2eλ2tv→2+...+cneλntv→n+b1eλ1tv→1+b2eλ2tv→2+...+bneλntv→nx→1(t)+x→2(t)=c1+b1eλ1tv→1+c2+b2eλ2tv→2+...+cn+bneλ2tv→n

Assume cn+bn=dn, substitute the values dnfor cn+bn, for all in the equation x→1(t)+x→2(t)=c1+b1eλ1tv→1+c2+b2eλ2tv→2+...+cn+bneλ2tv→nas follows:

x→1(t)+x→2(t)=c1+b1eλ1tv→1+c2+b2eλ2tv→2+...+cn+bneλ2tv→nx→1(t)+x→2(t)=d1eλ1tv→1+d2eλ2tv→2+...+dneλntv→nx→(t)=d1eλ1tv→1+d2eλ2tv→2+...+dneλntv→n

By the definition of solution of the linear system, x→t=x1→t+x2→tis a solution.

Hence, ifx→1(t) and x→2(t)is a solution of the system dx→dtAx→ thenx→t=x1→t+x2→t is a solution.

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