Chapter 6 Practice Test
6.1
Q: Find the general solution to the exact differential equation.
dy/dx = 5x^4 - 5 sec^2 x
A:
1. Find the anti-derivative: x^5 - 5 tan x
2. Take into consideration the constant
3. Answer: x^5 - 5 tan x + C
Q: Solve the initial value problem explicitly:
(1/1+t^2) + 3^t ln3 and y = 8 when t=0
A:
1. Find Anti-Derivative: Y = tan^(-1) t + 3^(t) + C
2. To solve explicitly, find C by using the given values for x and y:
(8) = tan^(-1) (0) + 3^(0) + C
7=C
3. Plug C into the initial value problem and you will have the answer :
y = tan^(-1)t + 3^t + 7
4) Find the general solution to the exact differential equation:
dy/du = 4(sin u)3(cos u)
Solution:
∫dy = ∫4 (sin u)3(cos u)du
y = (sin u)4 + C
5) Solve the initial value problem
dy/dx = 5sec2x - 3/2√x and y = 7 when x=0
Solution:
∫ dy/dx = ∫ 5sec2x - 3/2√x
y = 5tanx - x3/2 + C
7 = 5tan(0) - (0)3/2 + C
7=C
y = 5tanx - x3/2 + 7
6) Construct a slope field using the differential equation . Use slope analysis, not your
graphing calculator.
dy/dx = x- 2y
Solution:
6.2
U-substitution
Use Substitution to evaluate the integral
1. ∫√(cotx)(csc2 x dx
Let U = cot(x)
du= -csc2 x dx
∫√(cotx)(csc2 x dx= -∫u1/2du
= (-⅔)u3/2 + C
= -⅔(cotx)3/2+C
2.
5
6.3
Anti-Differentiation By Parts formula:
1) Integrate ∫ x3/(x+5)2dx
Worked out solution:
and
and
.
So that,
.
2) Integrate ∫x ⋅ cos(x/2)dx
Worked out solution:
2
3) Integrate using the tabular integration method: ∫ x4e-xdx
Begin by making two columns. On the left column, write the polynomial. Below it, write each
derivative until we reach zero. (That is why this one works for a polynomial function. No other
function will eventually reach zero.)
Now, on the right column, write the dv. Then below it list each anti-derivative. Continue until you
are lined up with the zero in the first column. For this example, the table would like the following:
f(x) and its derivatives
g(x) and its integrals
x4
e-x
4x3
-e-x
12x2
e-x
24x
-e-x
24
e-x
0
-e-x
Now, we go through and draw arrows diagonally down and to the right, starting at the top left.
And off to the left of the columns, we alternate writing a positive or negative sign, starting with a
positive sign. We continue in this manner until we are level with the zero in the first column. Now,
just multiply follow the arrows. Along each arrow, we multiply terms. Then add or subtract terms
depending on the sign.
=
4) Integrate ∫e-x⋅ cos(x)dx
6.4
Differential Equations and Mathematical Modeling
1) 1. (dy/dx) = (y)/(2√x) and y=1 when x=4
∫(dy/y)= (1/2) ∫x-1/2dx so ln abs(y) = √(x) + c. The initial point yields ln 1 =√(4) + c; hence c=-2.
With y > 0, the particular solution is ln(y) = √(x)-2 or y=e√(x) -2)
2. Solutions of the differential equation y dy=x dx are of the form…
Integrating yields (y2/2)=(x2/2) +C or y2= x2 + 2C or y2= x2+ C’, where we have replaced the
arbitrary constant 2C by C’ x2 -y2=C
3. Find the domain of the particular solution to the differential equation in Questions 2 that
passes through point (-2,1).
For initial point (-2,1), x2- y2 = 3. Rewriting the differential equation ydy=xdx as (dy/dx)= (x)/(y)
reveals that the derivative does not exist when y=0, which occurs at x=(+/-√(3). Since the
particular solution must be differentiable in an interval containing x=-2, the domain is x<-√(3)
4.If radium decomposes at a rate proportional to the amount present, then the amount R left
after t yr, if R sub zero is present initially and c is negative constant of proportionality, is given
by..
Since (dR/dt)= cR. (dR/R) = c dt, and ln ® = ct + C. When t=0, R=R sub zero; so ln (R sub
zero)=C or ln(R) = ct + ln (R subzero). Thus,
ln R - ln (R sub zero) = ct; ln (R/Rsubzero) = ct or (R/Rsubzero)= ect
5. The population of a city increases continuously at a rate proportional, at any time, to the
population at that time. The population doubles in 50 yr. After 75 yr the ratio of the population P
to the initial population P sub zero is given by….
The question given rise to the differntiaon equation (dP/dt)= kP, where P=2Psub zero, when
t=50. We seek P/Psub zero, for t=75. We get ln (P/Psub zero) = kt with ln (2)= 50k; then
ln (P/Psub zero)=(t/50)(ln 2) or (P/Psub zero)= 2t/50
6. A cup of coffee at temperature 180 degrees fahrenheit is placed on a table in a room at 98
degrees farenheit. The differential equation for its temperature at time t is (dy/dt)= -.11(y-68);
y(0)= 180. After 10 min the temperature (in degrees fahrenheit) of the coffee is...
We separate the variables in the given differential equation, then solve:
dy/(y-68)= -.11dt
ln (y-68) = -.11t+c
Since y(0)= 180, ln 112= c. Then
ln(y-58/112) = -.11t,
y=68 + 112e-.11t
When t=10, y=68+112e-1.1 which is about 105 degrees fahrenheit
7. Approximately, how long does it take the temperature of the coffee in question 6 to drop to
75 degrees fahrenheit…
The solution of the differential equation in questions 6, where y is the temperature of the coffee
at time t, is
y=68+112e-.11t
We find t when y=75 degrees fahrenheit:
75=68+112e-.11t
(7/112) = e-.11t
(ln 7 - ln 112)/(-.11) = t which is about 25 mins