The indicated function y1(x) is a solution of the given equation. Use reduction of order or this formula, y2 = y1(x) e−∫P(x) dx y12(x) dx , as instructed, to find a second solution y2(x). y'' − 4y' + 4y = 0; y1 = e2x y2 =

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'.

yes   because as the functions are graphed they eventually will intersect unless otherwise

step-by-step explanation:

step-by-step explanation:

for volume, add base x height x width