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Consider a linear system dx→dt=Ax→ of arbitrary size. Supposex→1(t) is a solution of the system and k is an arbitrary constant. Is x→t=kx→1ta solution as well? How do you know?

Short Answer

Expert verified

Yes, x→t=kx→1twould be the solution as well sincedkx→1dt=kdx→1dt .

Step by step solution

01

The matrix form of the derivatives

The rates of change of the components of the state vector x→t , or its derivatives dx1dt,dx2dt,…,dxndt.

When these rates depend linearly on x1,x2,…,xn, write the matrix equation asdx→dt=Ax→

.

02

Check whether x→t=kx→1tis a solution as well

When x→1twould be a solution of the linear system dy→dt=Ay→would satisfy this matrix equation.

Consider k as an arbitrary constant. Utilize the linearity of the differentiation as follow:

dx→dt=dkx→1dt=kdx→1dt=kAx→1=Akx→1=Ax→

Therefore,x→t=kx→1t would be a solution of the given system.

Thus, yes,x→t=kx→1t would be the solution as well sincedkx→1dt=kdx→1dt .

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