Solving a systemof eqns by given initial conditionsx′(t)=2x+8yy′(t)=−x−2yWhich has the I.C.&nbsp;X(0)=(6,−2). So I take that it means this:6=2x+8y−2=−x−2yand then it is solved as any other system? We get eigenvector&nbsp;(1,12), but since the eigenvalues of the system are two,&nbsp;±2i, we have to find the second generalized eigenvector.(2-2i8-1-2-2i)(112)=(6-2i-2-i)So the second vector would be&nbsp;(6−2i,−2−i). Using the general solutions for the system, we gety(t)=e2it(1,12)+e−2it(6−2i,−2−i)Would this be correct, or did I misinterpret the initial conditions?

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Solving a systemof eqns by given initial conditionsx′(t)=2x+8yy′(t)=−x−2yWhich has the I.C.&amp;nbsp;X(0)=(6,−2). So I take that it means this:6=2x+8y−2=−x−2yand then it is solved as any other system? We get eigenvector&amp;nbsp;(1,12), but since the eigenvalues of the system are two,&amp;nbsp;±2i, we have to find the second generalized eigenvector.(2-2i8-1-2-2i)(112)=(6-2i-2-i)So the second vector would be&amp;nbsp;(6−2i,−2−i). Using the general solutions for the system, we gety(t)=e2it(1,12)+e−2it(6−2i,−2−i)Would this be correct, or did I misinterpret the initial conditions?

Kailee Castro · Accepted Answer

Step 1If&amp;nbsp;A=28-1-2then&amp;nbsp;exp(tA)=sin(2t)+cos(2t)4sin(2t)-12sin(2t)cos(2t)-sin(2t).Step 2So, if&amp;nbsp;f(t)=exp⁡(tA).(6,−2), you have&amp;nbsp;f(0)=(6,−2)&amp;nbsp;and&amp;nbsp;f′(t)=A.f(t). So, take(x(t),y(t))=f(t)&amp;nbsp;=(6cos⁡(2t)−2sin⁡(2t),−sin⁡(2t)−2cos⁡(2t)).

undodaonePvopxl24 · Accepted Answer

Step 1In your case, the solution writesx(t)=a1eλ1tv1+a2eλ2tv2using the eigenvector/eigenvalue of the matrix A in the relation&amp;nbsp;x˙(t)=Ax(t).Both coefficients&amp;nbsp;a1,a2&amp;nbsp;are found using the provided initial conditionsx(0)=a1v1+a2v2which is again a 2-by-2 linear system..UPDATE With Matlab commands,A=[2,8;−1,−2];[U,D]=eig(A);U[6;−2]Step 2I found&amp;nbsp;ak=3.1820±1.0607i&amp;nbsp;and&amp;nbsp;v1=[0.9428;−0.2357+0.2357i]&amp;nbsp;Thusx(t)=a1e2itv1+a1⋅e−2itv1⋅=2R[a1e2itv1]UPDATE 2 Because eigenvalues/eigenvectors are complex conjugate, it holdsx(0)=a1v1+a1⋅v1⋅=2R(a1v1)Ax(0)=λ1a1v1+λ2a2v2=4iI(a1v1)Thus&amp;nbsp;2a1v1=x(0)−iAx(0)2From here simplifications occur&amp;nbsp;x(t)=R(2a1v1e2it)&amp;nbsp;=x(0)cos⁡(2t)+Ax(0)2sin⁡(2t)

Solving a systemof eqns by given initial conditions

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