Verify that the indicated function y = (x) is an explicit solution of the given first-order
differential equation. y? = 2xy^2; y = 1/(4 - x)^2 When y = 1/(4 - x)^2 Thus, in terms of x, 2xy^2
Since the left and right hand sides of the differential equation are equal when 1/(4 -x^2) is
substituted for y, y = 1/(4 -x)^2 is a solution. Proceed as in Example 4 of Section 1.1; by
considering simply as a function, give its domain. (Enter your answer using interval notation.)
Then by considering as a solution of the differential equation, give at least one interval I of
definition.
Solution
y = 1/(4 - x^2)
y\' = -1/(4 - x^2)^2 * (-2x)
y\' = 2x / (4 - x^2)^2 ---> first answer
Thus, in terms of x,
2xy^2 = y\'
2xy^2 = 2x / (4 - x^2)^2 ---> second answer
Here, phi(x) = 1/(4 - x^2)
Denominator cnanot equal 0
4 - x^2 = 0
x^2 = 4
x = 2 , -2
So, x cannot be 2 or -2
So, domain ---> (-inf , -2) U (-2 , 2) U (2 , inf) ---> third answer
Then by considering .......
(-2 , 2) ---> fourth answer.
Ecosystem Interactions Class Discussion Presentation in Blue Green Lined Styl...
Vertical Curyes_Examples - Pa FRESNGSTATE Lyles College of Engineerin.pdf
1. Vertical Curyes_Examples - Pa FRESNGSTATE Lyles College of Engineering VERTICAL
CURVES Given: 4 pvc g,--4.00% intersects g2-+3.80% at PVI of 52 + 00 with elevP,-1261.50,
with L = ? feet. Design an equal tangent parabolic curve to meet a railroad crossing, which exists
at sta 5350 with an elev of 1271.20' 2 3.80 + 4.00 2L 271.20 1261.50+4.00 0.97512 5=0 2a 9.1
152 sta 1 10:54/11:00
Solution
BVC: Beginning of Vertical Curve aka PVC
V: Vertex aka PVI
EVC: End of Vertical Curve aka PVT
g1: percent grade of back tangent
g2: percent grade of forward tangent
L: curve length (horizontal distance) in feet or stations
x: horizontal distance from any point on the curve to the BVC
r: rate of change of grade
r = (g2 – g1)/L
where:
g2 & g1 - in percent (%)
L – in stations
and
Y = YBVC + g1x + (r/2)x2
where:
YBVC – elevation of the BVC in feet
L = STAEVC – STABVc
L = STAEVC – STABVC
L = 1237.56 – 1230 = 7.56 x 60 /100
= 4.5
r = (g2 – g1) / L
r = (-3.5 – 2) / 4.5
r = 1.22
r/2 = 0.611Stationx (stations)g1xr/2 x2Curve Elevation20 + 25.32 BVC0001222.12520 +
500.24680.8638-0.0371222.95221 + 501.24684.3638-0.9481226.36722 + 502.24687.8638-
3.0791231.15223 + 503.246811.3638-6.4301236.085
