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Diff Eq 1.5

 

<== Differential Equations
Number Question Answer
See the work
1. Find the particular solution of the differential equation
dy/dx+3y=6
satisfying the initial condition y(0)=0.
Answer: y= .
Your answer should be a function of x.
((2e^(3x))-2)/e^(3x)
The work

Point Cost: 3
2. Find the particular solution of the differential equation
dy/dx+ycos(x)=6cos(x)
satisfying the initial condition y(0)=8.
Answer: y(x)= .
6+2*(e^[-sin(x)])
The work

Point Cost: 3
3. Find the general solution, y(t), which solves the problem below, by the method of integrating factors.
7t(dy/dt)+y=t^4

Put the problem in standard form.
Then find the integrating factor, 
ho(t) = ,
and finally find y(t)= . (use C as the unkown constant.)

p(t)=t^(1/7)

y(t)=t^4/29+Ct^(-1/7)

The work

Point Cost: 3
4. Solve the following initial value problem:
t(dy/dt)+6y=6t
with y(1)=7
Put the problem in standard form.
Then find the integrating factor, 
ho(t) = ,
and finally find y(t)= .
p(t)=t^6
y(t)=((6/7)t^7+(43/7))/(t^6)
The work

Point Cost: 3
5. Solve the initial value problem
9(sin(t)+cos(t)y)=cos(t)sin^2(t),

for 0<t<π and y(π/2)=16.
Put the problem in standard form.
Then find the integrating factor, 
ho(t) = ,
and finally find y(t)= .
p(t)=sint
y(t)=((sin^3(t)+431)/27)/sin(t)
The work

Point Cost: 3
6. Solve the initial value problem
12(t+1)(dy/dt)-9y=27t,

for t > -1 with y(0)=20.
Put the problem in standard form.
Then find the integrating factor, 
ho(t) = ,
and finally find
y(t)= .
p(t)=(t+1)^-(9/12)

y(t)=((3(3t+4)/(t+1)^(3/4))+8)(t+1)^(9/12)
The work

Point Cost: 3
7. Solve the following initial value problem:
dy/dt+0.9ty=4t
with y(0)=5.
y= .
y=(40/9*e^(.45t^2)+5/9)/(e^(.45t^2))
The work

Point Cost: 3
8. Find the function y(t) that satisfies the differential equation
dy/dt-2ty=15(t^2)*e^(t^2)
and the condition y(0)=-1.
y(t)= .
(15/3*t^3-1)(e^(t^2))
The work

Point Cost: 3
9. Find the general solution to the differential equation
x^2+1xy+x(dy/dx)=0
Put the problem in standard form.

Find the integrating factor, 
ho(x) = .

Find y(x)= .
Use C as the unknown constant.
p(t)=e^x
y(x)=-x+1+(C/e^x)
The work

Point Cost: 3
10. Solve the initial value problem
dy/dt-y=8e^t+35e^(6t)
with y(0)=2
y= .
y=((8t+7e^(5t))-5)e^t
The work

Point Cost: 3
11. A tank contains 2760 L of pure water. Solution that contains 0.02 kg of sugar per liter enters the tank at the rate 5 L/min, and is thoroughly mixed into it. The new solution drains out of the tank at the same rate.
(a) How much sugar is in the tank at the begining?
y(0)= (kg)

(b) Find the amount of sugar after t minutes.
y(t)= (kg)

(c) As t becomes large, what value is y(t) approaching ? In other words, calculate the following limit. \displaystyle  \lim_{t 	o \infty} y(t) = (kg)
y(0)=0
y(t)=2760*.02*(1-e^(-5/2760*t))

\displaystyle  \lim_{t 	o \infty} y(t) =(2760)*(.02)
The work
12. A tank contains 80 kg of salt and 1000 L of water. A solution of a concentration 0.04 kg of salt per liter enters a tank at the rate 7 L/min. The solution is mixed and drains from the tank at the same rate.

(a) What is the concentration of our solution in the tank initially?
concentration = (kg/L)

(b) Find the amount of salt in the tank after 2 hours.
amount = (kg)

(c) Find the concentration of salt in the solution in the tank as time approaches infinity.
concentration = (kg/L)
a) 0.08 kg/L

b)
80-40(1-e^((-7/1000)(120)))
c) 0.04
The work
13. A tank contains 50 kg of salt and 2000 L of water. Pure water enters a tank at the rate 12 L/min. The solution is mixed and drains from the tank at the rate 6 L/min.

(a) What is the amount of salt in the tank initially?
amount = (kg)

(b) Find the amount of salt in the tank after 4.5 hours.
amount = (kg)

(c) Find the concentration of salt in the solution in the tank as time approaches infinity. (Assume your tank is large enough to hold all the solution.)
concentration = (kg/L)
a) 50
b) ?
c) 0
The work