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Let Abe ann×nmatrix andK a scalar. Consider the following two systems:

dx→dt=Ax→(I)dc→dt=kAc→(II)

Show that ifx→(t)is a solution of the system(I)thenc→(t)=x→(kt)is a solution of the system(II). Compare the vector field of the two system.

Short Answer

Expert verified

Yes, c→t=x→ktis a solution of the systemdc→dt=kAc→

Step by step solution

01

Determine the equation of the solution

Consider x→tandc→tis a solution of the system dx→dt=Ax→anddc→dt=kAc→respectively.

Assume λ1,λ2,λ3,...,λnbe the Eigen values of the matrix A then there exist Eigen vectors v→1,v→2,v→3,...,v→nsuch that x→t=c1eλ1tv→1+c2eλ2tv→2+...+cneλntv→nand c→t=c1ekλ1tv→1+c2ekλ2tv→2+...+cnekλntv→nwhere c1,c2,...,cnis constant.

If xtis the solution of the linear system dx→dt=Ax→ then xt=c1eλ1tv→1+c2eλ2tv→2+...+cneλntv→nwhere λ1,λ2,λ3,...,λnbe the Eigen values and v→1,v→2,v→3,...,v→nof n×nmatrix A.

02

Show that c→t=ektx→t is a solution of the system 

Simplify the equation c→t=c1ekλ1tv→1+c2ekλ2tv→2+...+cnekλntv→nas follows.

c→t=c1ekλ1tv→1+c2ekλ2tv→2+...+cnekλntv→nc→t=c1eλ1ktv→1+c2eλ2ktv→2+...+cneλnktv→n

As x→t=c1eλ1tv→1+c2eλ2tv→2+...+cneλntv→n, substitute the value ktfor t in the equation x→t=c1eλ1tv→1+c2eλ2tv→2+...+cneλntv→nas follows.

x→t=c1eλ1tv→1+c2eλ2tv→2+...+cneλntv→nx→kt=c1eλ1ktv→1+c2eλ2ktv→2+...+cneλnktv→n

Substitute the value x→ktforc1eλ1ktv→1+c2eλ2ktv→2+...+cneλnktv→nin the equation c→t=c1eλ1ktv→1+c2eλ2ktv→2+...+cneλnktv→nas follows.

localid="1659535938530" c→t=c1eλ1ktv→1+c2eλ2ktv→2+...+cneλnktv→nc→t=x→kt

Hence, if x→tandc→tis a solution of the systemdx→dt=Ax→and dc→dt=kAc→respectively thenc→t=x→kt is a solution

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