Review for Test 1

TEST 1 - ODE REVIEW
The solutions are at the end. I suggest you try to solve the problems first and, then, check your
solution. That way you will get some practice in determining what type of equation you are facing
and which method to use. Also note that the same solution can be written in several different
ways. So if your solution does not coincide with mine, that does not necessarily mean that yours
is wrong; it could be just a few simple algebraic steps away from mine.
1) Use Euler’s method to approximate the solution to the IVP at x = 0.1, 0.2, 0.3, 0.4 and 0.5 using
a step size of 0.1 (round to three decimal places).
dy x
= ,
y(0) = −1
dx y
After obtaining the approximate values, proceed to solve the equation. On what interval, if any, is
the solution positive?
2) Solve the IVP
y0 (x) =
sec2 (y)
,
1 + x2
y(0) = 0
3) Solve the IVP and determine the interval of existence of the solution.
y0 (x) = 8x3 e−2y ,
y(1) = 0
4) Solve the IVP and determine the interval of existence of the solution.
1
dy
y(1) = −
y3 − θy2 = −2y θ2 ,
dθ
2
5) Find the general solution to
dy
3x2 − y2 + xy − x3 y−1
=0
dx
6) Solve the IVP using the integrating factor method outlined in class for first-order linear equations, and determine the interval of existence of the solution.
t2 + 1 y0 (t) = yt − y,
y(0) = 1
Note that the equation is also separable. Try solving it by separating the variables instead and
compare the two methods; which method is simpler in this case?
7) Find the general solution to
(t + y + 1) −
dy
=0
dt
1
2
8) Solve the IVP and determine the interval of existence of the solution.
2xy2 + 1
dx + dy = 0,
2x2 y
y(1) = 1
9) Find the general solution to
(ye xy + 2x) dx + (xe xy − 2y) dy = 0
10) Use your ingenuity1 to solve
1
dy
= 4y
dx e + 2x
Hint: The roles of the independent and dependent variables may be switched.
11) a) Consider a tank holding 1000 L of pure water into which a brine solution of salt begins to
flow at a constant rate of 6 L/min. The solution inside the tank is kept well-stirred and is flowing
out of the tank at a rate of 6 L/min. If the concentration of salt in the brine entering the tank is 0.1
Kg/L, determine when the concentration of salt in the tank will reach 0.05 Kg/L (Let A(t) denote
the amount (in Kg) of salt in the tank at time t ≥ 0).
b) Assume now that the brine leaves the tank at a rate of 5 L/min instead of 6 L/min, with all else
being the same. What is now the ODE that describes the evolution of A(t)? No need (but feel free)
to solve this last ODE; just write it down.
Solutions.
1) Below are the (xn , yn ) ordered pairs for n = 0, 1, 2, 3 and 4.
(0.1, −1), (0.2, −1.010), (0.3, −1.030), (0.4, −1.059), (0.5, −1.097).
The solution (obtained via the separable method) is
√
y = − x2 + 1
Clearly, y is negative for all x ∈ R. So y is never positive.
2) The equation may be solved via the separable method. The solution is implicitly defined by the
equation:
2y + sin 2y = 4 arctan x
3) The equation may be solved via the separable method. The solution and its interval of existence
are:
√ 1/2 √ 1/2
√
−∞, − 3/2
∪
3/2
, +∞
y = ln 4x4 − 3,
4) The equation is homogeneous. The solution and its interval of existence are:
θ
(−∞, +∞)
y=−
√ ,
1 + |θ|
1
Ingenuity in this case does not refer to computer software.
3
5) The equation is homogeneous. The general solution is:
ln y2 /x6 − (y/x)2 = C
6) The equation is linear (also separable). We solve it by using the integrating factor method. The
solution and its interval of existence are:
√
1 + t2
(−∞, +∞)
y = arctan t ,
e
7) The equation is linear and may be solved using the integrating factor method. The general
solution is:
y = −t − 2 + Cet
8) This equation is exact. The solution and its interval of existence are:
√
2−x
(−∞, 0)∪(0, 2]
y=
,
|x|
9) This equation is exact. The general solution is defined implicitly by the equation:
e xy + x2 − y2 = C.
10) Note that the equation is nonlinear under the usual convention that y represents the dependent
variable and x the independent one; however, if we reverse the roles of these variables and flip the
equation to get
dx
dx
= e4y +2x,
or equivalently
−2x = e4y ,
dy
dy
then, with x the dependent variable and y the independent one, the above equation is now linear.
Using the integrating factor
R
µ(y) = e− 2 dy = e−2y ,
we obtain the general solution
x(y) =
e4y
+ Ce2y
2
11) a) The IVP describing the evolution of A(t) is
dA
3A
= 0.6 −
,
A(0) = 0,
dt
500
which is both linear and separable. Solving yields
A(t) = 100 1 − e−3t/500 .
Consequently, the concentration of salt in the tank at time t ≥ 0 is
A(t)
= 0.1 1 − e−3t/500 Kg/L.
1000
To determine the time it takes for the concentration to reach 0.05 Kg/L, we then solve
0.05 = 0.1 1 − e−3t/500
4
for t, and find
t=
500
ln 2 ∼ 115.52 min.
3
b) Now the brine leaves the tank at a rate of 5 L/min and enters it at a rate of 6L/min. This means
that the brine solution in the tank (recall the tank holds 1000 L of pure water at t = 0) increases by
1L every minute. As a result, the rate of flow of A out of the tank will be2
A(t)
5A
(5L/min) ×
Kg/L =
Kg/min.
1000 + t
1000 + t
The term 1000 + t is just the straight line that describes how the brine solution in the tank increases
(linearly). It connects the point (0, 1000) (1000 L at t = 0 min.) to, for instance, the point (10, 1010)
(1010 L at t = 10 min.). The IVP for A(t) then becomes
dA
5A
= 0.6 −
,
A(0) = 0.
dt
1000 + t
2
The rate in is still the same as before because only the rate at which the brine exits the tank has changed.

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