#$&*
course Mth 174
10/3 9am
Section 6.31) 6.3.2 Find the general solution of the differential equation dy/dx = 9x - 2/x.
A: Finding the general solution for a differential equation is like finding the anti derivative. The anti derivative of 9x is (9/2)x^2 and bringing the -2 out, we see that the anti derivative of (1/x) is ln(x). The general solution is y=(9/2)x^2 - 2ln(x) +c.
2) 6.3.4 Find the general solution of the differential equation dr/dp = 4 cos p.
A: The general solution is r=4 sin p + c.
3) 6.3.6 Find the solution of the initial value problem: dy/dx = 8x^3 +2x, y(2) = 10.
A: Taking the anti derivative, we get y=2x^4 + x^2 +c.
Substituting the initial value for y and x we get
10=2(2)^4 + (2)^2 +c. which works out to be c=-26.
So, the solution of the initial value problem is
y=2x^4 + x^2 - 26.
4) 6.3.8 Find the solution of the initial value problem: ds/dt = -32t + 100, s = 50 when t = 0.
A: The general solution is s= -16t^2 + 100t + c. When s=50 and t=0, we find that c=50.
So, the solution to the initial value problem is s= -16t^2 + 100t + 50.
5) 6.3.10 Find the general solution of the differential equation dy/dx = 2x + 1. Graph three different solutions of this equation.
A: The general solution is y= x^2 + x + c.
Picking a solution is equivalent to picking any value of c. To do this we have to pick initial values.
I’ll pick the points (0,0), (0,1), and (1,0).
Which gives us c values of 0, 1, and -2.
Which gives us the equations
y= x^2 + x + 0
y= x^2 + x + 1
y= x^2 + x + -2
The graphs are concave up parabolas that cross the y axis at the given c values and also go through the chosen x and y values.
6) 6.3.16 A water balloon is launched from the roof of a building at time t = 0 and velocity v(t) = -32t + 40ft/s at time t. v > 0 corresponds to vertical motion. If the roof of the building is 30 feet above the ground, find an expression for the height of the water balloon above the ground at time t. What is the average velocity of the balloon between t = 1.5 and t = 3 seconds? For a 6-foot person is standing on the ground below, how fast is the balloon when it hits them in the head.
A: Taking the anti derivative of the velocity equation, we find that the distance equation is s=-16t^2 +40t +c.
Given that s=30 when t=0, we see that c=30 so
the expression for the height of the water balloon above the ground at time t is s=-16t^2 + 40t + 30.
Plugging in the times t=1.5 and t=3 into the velocity equation, we see that the average velocity is -32ft/s.
When s=6, we get the quadratic form 2t^2 -5t -3=0
giving us t=3.
Plugging t=3 into the velocity equation, we find that the water balloon was traveling -56ft/s when it struck the person in the head."
@& When problems are submitted in this format, I can't guarantee that every detail of your arithmetic checks out. However I have read your solutions and checked out most of the details, and see no problem here. You are following the appropriate steps on each problem and other than minor errors I might have missed, your work appears to be correct.*@