Respuesta :
Answer:
The general solution of given system is [tex]x(t)=C_1e^{6t}\cos t+C_2e^{6t}\sin t[/tex] and [tex]y(t)=C_1e^{6t}\cos t+C_2e^{6t}\sin t[/tex].
Step-by-step explanation:
The given differential equations are
[tex]\frac{dx}{dt}=8x-y[/tex] .... (1)
[tex]\frac{dy}{dt}=5x+4y[/tex] .... (2)
Differentiate both equations with respect to t.
[tex]x''=8x'-y'[/tex] ...... (3)
[tex]y''=5x'+4y'[/tex] ..... (4)
Using equation (3) and (2).
[tex]x''=8x'-(5x+4y)[/tex]
[tex]x''=8x'-5x-4y)[/tex]
[tex]x''=8x'-5x-4(8x-x')[/tex] (Using 1)
[tex]x''=8x'-5x-32x+4x')[/tex]
[tex]x''=12x'-37x[/tex]
[tex]x''-12x'+37x=0[/tex]
[tex]r^2-12r+37=0[/tex]
Using Quadratic formula
[tex]r=\frac{-b\pm \sqrt{b^2-4ac} }{2a}[/tex]
[tex]r=\frac{12\pm \sqrt{144-148} }{2(1)}[/tex]
[tex]r=6\pm i[/tex]
If [tex]r=\alpha+i\beta[/tex], then
[tex]y(x)=e^{\alpha x}(C_1\cos \beta x+C_2\sin \beta x)[/tex]
Therefore x(t) is defined as
[tex]x(t)=e^{6t}(C_1\cos t+C_2\sin t)[/tex]
Using equation (4) and (1).
[tex]y''=5(8x-y)+4y'[/tex]
[tex]y''=40x-5y+4y'[/tex]
[tex]y''=40(\frac{y'-4y}{5})-5y+4y'[/tex] (Using 2)
[tex]y''=8(y'-4y)-5y+4y'[/tex]
[tex]y''=8y'-32y-5y+4y'[/tex]
[tex]y''+12y'-37y=0[/tex]
[tex]r^2-12r+37=0[/tex]
Using Quadratic formula
[tex]r=\frac{-b\pm \sqrt{b^2-4ac} }{2a}[/tex]
[tex]r=\frac{12\pm \sqrt{144-148} }{2(1)}[/tex]
[tex]r=6\pm i[/tex]
Therefore y(t) is defined as
[tex]y(t)=e^{6t}(C_1\cos t+C_2\sin t)[/tex]
