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Let be an matrix anda scalar. Consider the following two systems:

dx→dt=Ax→(l)dc→dt=(A+kln)c→(ll)

Show that if x→(t) is a solution of the system (l)then role="math" localid="1659701582223" c→(t)=ektx→(t)is a solution of the system (ll).

Short Answer

Expert verified

Yes,c→t=ektx→t is a solution of the system dc→dt=(A+kln)c→.

Step by step solution

01

Determine the equation of the solution.

Considerx→tandc→tis a solution of the systemdx→dt=Ax→anddc→dt=A+klnc→respectively.

Assumeλ1,λ2,λ3,…,λnbe the Eigen values of the matrix A then there exist Eigen vectorsv→1,v→2.v→3,…,v→nsuch that x→1(t)=c1eλ1tv→1+c2eλ2tv→2+...+cneλntv→nandc→(t)=c1eλ1+kv→1+c2eλ2+kv→2+...+cneλn+kv→nwherec1,c2,…,cnis constant.

If x(t) is the solution of the linear system localid="1659702528208" dx→dt=Ax→thenx→1(t)=c1eλ1tv→1+c2eλ2tv→2+...+cneλntv→nwhereλ1,λ2,λ3,…,λnbe the Eigen values and ofn×nmatrix A.

02

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

Simplify the equationx→1(t)=c1eλ1tv→1+c2eλ2tv→2+...+cneλntv→nas follows.

c→(t)=c1eλ1+kv→1+c2eλ2+kv→2+...+cneλn+kv→nc→(t)=c1eλ1tektv→1+c2eλ2tektv→2+...+cneλntektv→nc→(t)=ektc1eλ1tv→1+c2eλ2tv→2+...+cneλntv→nc→(t)=ektx→t

By the definition of solution of the linear system,c→t=ektx→tis a solution.

Hence, if x→tandc→tis a solution of the systemdx→dt=Ax→anddc→dt=A+klnc→respectively then is a solution.

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