Respuesta :
Answer:
The answer is "[tex]\bold{e^{-t} \sin t}[/tex]"
Explanation:
Given equation:
[tex]\to x+2x+2x=0.............(a)[/tex]
Let x= e^{mt} be a solution of the equation will be:
[tex]\to m^2+2m+2=0[/tex]
compare the value with the [tex]am^2+bm+c=0[/tex]
[tex]\to a= 1\\\to b= -2\\\to c= 2\\[/tex]
Formula:
[tex]\bold{=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\\\\=\frac{-2 \pm \sqrt{(-2)^2-4\times 1 \times 2}}{2 \times 1}\\\\=\frac{-2 \pm \sqrt{4-8}}{2}\\\\=\frac{-2 \pm \sqrt{-4}}{2}\\\\=\frac{-2 \pm 2i}{2}\\\\[/tex]
Calculate the g.s at equation (a):
[tex]\to x = e^{-t} (c_1 \cos t+ c_2 \sin t)............(b)\\\\\to x = e^{-t} (c_2 \cos t -c_1 \sin t) - e^{-t} (c_1 \cos t + c_2\sin t) ...............(c)\\\\[/tex]
[tex]_{when} \\\\\to t= 0\\\\\to x=x_0\\\\\to x= v_0 =1 \\\\ \ then \ form \ equation \ (b) \ and \ equation \ (c)[/tex]
[tex]\to a= 0 \\ \to c_2 -c_1 =1 \\ \to c_2=1[/tex]
The value of [tex]x = e^{-t} ( 0 \times \cos t + 1 \times \sin t)[/tex]
[tex]\boxed{\bold{x = e^{-t} (\sin t)}}[/tex]
