Asystem of two differential equations with real-valued coefficients has the complex solution el1+)t(- 1 2 i 2 2 i then two real-valued solutions are given by -cos (t) and e\2 cos (t)) a) -2 sin (t) e -2 sin (t) 2 sin (t)- cos (t) ,) b) and 2 sin (t)+2 cos (t) e -sin (t)- 2 cos (t) 2 sin (t)- 2 cos (t)
-2 sin (t) - cos () | and 2 cos (t)-2 sin (t) c) e -sin (t) 2 cos (t) 2 sin (t)-2 cos (t) 2 i sin (i t)-cos (i t) and 2 i sin (i t)+2 cos (i t) d) -i sin (i t) - 2 cos (i t) 2 i sin (i t)-2 cos (i t).

ab: 2 n83b8eb7ebi. 8e yen in

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.