Find the solution of the sistemx″=2x+y and y″=x+2y

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Vegljamzt6 · Accepted Answer

Step 1Let x1=x, x2=x′, x3=y, x4=y′You can check that the given second-order system is equivalent to the following first-order system:x1′=x2x2′=2x1+x3x3′=x4x4′=x1+2x3Define the vector function x=(x1,x2,x3,x4)T. We have that x′=Ax, where the coefficient matrix isA=[0100201000011020].WolframAlpha gives the eigenvalues of A as λ=±1,±3 which are distinct and so the general solution isx(t)=c1etv1+c2e−tv2+c1e3tv3+c4e−3tv4.Step 2One can also solve the second-order linear system using the eigenvalue method without rewriting it as an equivalent first-order system. Let z=(x,y)T. You can check that z=eαtv is a solution to z″=Bz with α2=λ and (λ,v) an eigenpair of the matrix B, i.e., Bv=λv. In this case,B=[2112]with distinct eigenvalues λ=1,3 and so the general solution isz(t)=c1etw1+c2e−tw1+c3e3tw2+c4e−3tw2.

annieljcddj0 · Accepted Answer

Step 1Notice that(x−y)″=x−y.Set u=x−y then you have the following IVPu″=u,u(0)=0,u′(0)=2.After you solve for u then you see that y=x−u which means x″=2x+x−u=3x−u where u is known.

Find the solution of the sistem \(\displaystyle{x}{''}={2}{x}+{y}\) and

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