Find the general solution of the system x' = 4x + y, y' = -8x + 8y.

Question

Find the general solution of the system

[latex]\begin{aligned}&x^{\prime}=4 x+y, \&y^{\prime}=-8 x+8 y .\quad\quad\quad\quad\quad (12)\end{aligned}[/latex]

Accepted Answer

Here

[latex]D=\left|\begin{array}{cc}4-\lambda & 1 \-8 & 8-\lambda\end{array} ight|=\lambda^{2}-12 \lambda+40=0[/latex].

The roots of the characteristic equation are [latex]\lambda_{1}=6+2 i[/latex] and [latex]\lambda_{2}=6-2 i[/latex], so that Theorem 3 yields the linearly independent solutions

[latex]\begin{aligned}&\left(x_{1}(t), y_{1}(t) ight)=\left(e^{a t}\left(A_{1} \cos b t-A_{2} \sin b t ight), e^{a t}\left(B_{1} \cos b t-B_{2} \sin b t ight) ight) \&\left(x_{2}(t), y_{2}(t) ight)=\left(e^{a t}\left(A_{1} \sin b t+A_{2} \cos b t ight), e^{a t}\left(B_{1} \sin b t+B_{2} \cos b t ight) ight)\end{aligned}[/latex]

[latex]\begin{aligned}&\left(x_{1}(t), y_{1}(t) ight)=\left(e^{6 t}\left(A_{1} \cos 2 t-A_{2} \sin 2 t ight), e^{6 t}\left(B_{1} \cos 2 t-B_{2} \sin 2 t ight) ight) \&\left(x_{2}(t), y_{2}(t) ight)=\left(e^{6 t}\left(A_{1} \sin 2 t+A_{2} \cos 2 t ight), e^{6 t}\left(B_{1} \sin 2 t-B_{2} \cos 2 t ight) ight)\end{aligned}[/latex]

Substituting the first equation into system (12) yields, after a great deal of algebra, the system of equations

[latex]\begin{aligned}\left(2 A_{1}-2 A_{2}-B_{1} ight) \cos 2 t-\left(2 A_{1}+2 A_{2}-B_{2} ight) \sin 2 t &=0 \\left(8 A_{1}-2 B_{1}-2 B_{2} ight) \cos 2 t-\left(8 A_{2}+2 B_{1}-2 B_{2} ight) \sin 2 t &=0\end{aligned}[/latex]

Since [latex]t[/latex] is arbitrary and the functions [latex]\sin 2 t[/latex] and [latex]\cos 2 t[/latex] are linearly independent, the terms in parentheses must all vanish. A choice of values that will accomplish this is [latex]A_{1}=1, A_{2}=\frac{1}{2}, B_{1}=1[/latex], and [latex]B_{2}=3[/latex]. Thus two linearly independent solutions to system (12) are [latex]\left(e^{6 t}\left(\cos 2 t-\frac{1}{2} \sin 2 t ight), e^{6 t}(\cos 2 t-3 \sin 2 t) ight)[/latex] and [latex]\left(e^{6 t}\left(\sin 2 t+\frac{1}{2} \cos 2 t ight), ight.[/latex], [latex]\left.e^{6 t}(\sin 2 t+3 \cos 2 t) ight)[/latex]. The general solution of system (12) is a linear combination of these two solutions.

Question 12.3.3: Find the general solution of the system x' = 4x + y, y' = -8...

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