Download Math 244, Quiz 1 Solutions 1. Verify that y(x) = Ce −x +x−1 satisfies

Math 244, Quiz 1 Solutions
1. Verify that y(x) = Ce−x + x − 1 satisfies the differential equation y 0 = x − y. (Do not solve the equation, verify
by substitution.) Then determine the value of C so that y(x) satisfies the initial condition y(0) = 4.
Substitute y = Ce−x + x − 1 and y 0 = −Ce−x + 1 into the differential equation to obtain
−Ce−x + 1 = x − (Ce−x + x − 1) .
This simplifies to
−Ce−x + 1 = −Ce−x + 1 ,
which is true for all x.
To determine the constant C when y(0) = 4, substitute x = 0 and y(0) = 4 into the solution formula to obtain
4 = Ce0 + 0 − 1 .
Consequently, C = 5.
2. Do part a or part b, not both.
a) Find the solution to the initial value problem xy 0 = y + x, y(1) = π . Show each step in the calculation.
Write the differential equation in the form xy 0 − y = x to see that it is first order linear. Normal form is
y0 −
R
The integrating factor is ρ = e
−1/x dx
1
y = 1.
x
(1)
= e− ln(x) = 1/x. Multiply equation (1) by 1/x and simplify to
0
1
1
y = .
(2)
x
x
Integrate both sides and use the fact that x > 0 to obtain
1
y = ln(x) + C ,
x
and
y = x ln(x) + Cx .
This is the general solution when x > 0.
To determine the solution satisfying the initial condition y(1) = π substitute x = 1 and y = π to get the
equation π = C so the solution to the initial value problem is y = x ln(x) + πx.
b) Find the solution to the initial value problem y 0 = y sin(x), y(π) = 1 . Show each step in the calculation.
This is a separable differential equation: dy/y = sin(x) dx. Integrate to obtain ln(y) = − cos(x) + C. This
is the general solution provided y > 0.
The value of the constant C is obtained by substituting x = π and y = 1 into the solution formula. Doing
so yields 0 = 1 + C so C = −1 and
ln(y) = − cos(x) − 1 .
This implies that y = e− cos(x)−1 , or y = e− cos(x) /e.
3. At noon a car starts from rest at a point A and proceeds with constant acceleration along a straight road
toward point C, 35 miles away. If the constantly accelerated car arrives at C with a velocity of 60 mi/hr, at
what time does it arrive at C? What is its average velocity for the trip?
Let a denote the (constant) acceleration of the car, v its velocity, and x its position (as measured from the
starting point when time t = 0: x(0) = 0, v(0) = 0). Since v 0 = a, v(t) = at + v0 . The constant is 0
because v(0) = 0 and v(t) = at. Integrating again, x(t) = at2 /2 + x0 where, once more, the constant is 0 and
x(t) = at2 /2.
Let T denote the time it takes the car to reach point C. Then 35 = aT 2 /2 and 60 = aT . Divide the first
equation by the second one to see that 35/60 = T /2 so T = 70/60 = 7/6 hours. The car arrives at point C in
1 + 1/6 hours which is one hour and ten minutes, i.e. at 1:10 PM. The average velocity is distance (35 miles)
divided by time (7/6 hours). This is
35
= 30 miles per hour .
7/6