In each of problems, determine intervals in which solutions are sure to exist.

1. y^(4) + 4y"' + 3y = t
2. ty"' + (sin t)y" + 3y = cos t
3. t(t - 1)y^(4) + e^ty" + 4t^2y = 0
4. y"' + ty" + t^2y' + t^3y = ln t
5. (x-1)y^(4) + (x + 1)y" + (tan x)y = 0
6. (x^2 - 4)y^(6) + x^2y"' + 9y = 0

eveidence i thing that

step-by-step explanation:

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'.