Find the general solution of the system x' = -4x - y, y' = x - 2y.

Question

Find the general solution of the system

[latex]\begin{aligned}&x^{\prime}=-4 x-y, \&y^{\prime}=x-2 y .\quad\quad\quad\quad\quad (7)\end{aligned}[/latex]

Accepted Answer

Equation (3) is

[latex]\begin{aligned}D &=\left|\begin{array}{ll}a_{11}-\lambda & a_{12} \a_{21} & a_{22}-\lambda\end{array} ight| \&=\left(a_{11}-\lambda ight)\left(a_{22}-\lambda ight)-a_{21} a_{12}\end{aligned}[/latex]

[latex]D=\left|\begin{array}{cc}-4-\lambda & -1 \1 & -2-\lambda\end{array} ight|=(\lambda+4)(\lambda+2)+1=\lambda^{2}+6 \lambda+9=0[/latex]

which has the double root [latex]\lambda_{1}=\lambda_{2}=-3[/latex]. From system (2), with [latex]\lambda=-3[/latex], we find that

[latex]\begin{aligned}&\left(a_{11}-\lambda ight) \alpha+a_{12} \beta=0 \&a_{21} \alpha+\left(a_{22}-\lambda ight) \beta=0\end{aligned}[/latex]

[latex]\begin{array}{r}-\alpha_{1}-\beta_{1}=0, \\alpha_{1}+\beta_{1}=0 .\end{array}[/latex]

A nontrivial solution is [latex]\alpha_{1}=1, \beta_{1}=-1[/latex], yielding the solution [latex]\left(e^{-31},-e^{-31} ight)[/latex].
If we try to find a solution of the form [latex]\left(\alpha_{2} t e^{-3 t}, \beta_{2} t e^{-3 t} ight)[/latex] we immediately run into trouble, since the derivatives on the left-hand side of system (7) produce terms of the form [latex]c e^{-3 t}[/latex] not present on the right-hand side of (7). This explains why we must seek a solution of the form

[latex]\left(\left(\alpha_{2}+\alpha_{3} t ight) e^{-3 t},\left(\beta_{2}+\beta_{3} t ight) e^{-3 t} ight)\quad\quad\quad\quad\quad (8)[/latex].

Substituting the pair (8) into system (7), we obtain

[latex]\begin{aligned}&e^{-3 t}\left(\alpha_{3}-3 \alpha_{2}-3 \alpha_{3} t ight)=-4\left(\alpha_{2}+\alpha_{3} t ight) e^{-3 t}-\left(\beta_{2}+\beta_{3} t ight) e^{-3 t} \&e^{-3 t}\left(\beta_{3}-3 \beta_{2}-3 \beta_{3} t ight)=\left(\alpha_{2}+\alpha_{3} t ight) e^{-3 t}-2\left(\beta_{2}+\beta_{3} t ight) e^{-3 t}\end{aligned}[/latex]

Dividing by [latex]e^{-3 t}[/latex] and equating constant terms and coefficients of [latex]t[/latex], we obtain the system of equations

[latex]\begin{aligned}\alpha_{3}-3 \alpha_{2} &=-4 \alpha_{2}-\beta_{2} \-3 \alpha_{3} &=-4 \alpha_{3}-\beta_{3} \\beta_{3}-3 \beta_{2} &=\alpha_{2}-2 \beta_{2} \-3 \beta_{3} &=\alpha_{3}-2 \beta_{3}\end{aligned}[/latex]

One solution is [latex]\alpha_{2}=1, \beta_{2}=-2, \alpha_{3}=1, \beta_{3}=-1[/latex]. Thus a second solution of system (7) is [latex]\left((1+t) e^{-3 t},(-2-t) e^{-3 t} ight)[/latex]. It is easy to verify that the two solutions are linearly independent, since [latex]W(t)=-e^{-6 t}[/latex].

Question 12.3.2: Find the general solution of the system x' = -4x - y, y' = x...

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