Exam I Review: Solutions
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Classification
Classify the following initial value problems: state the order of each equation, whether they are linear
or non-linear, whether they separable, and classify the coefficients. Are any of them Bernoulli?
1. (yy 0 )0 + x2 y = 5; y(0) = 0
=⇒ Second order, non-linear, non-constant coeff’s
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2. y 00 + x−1/2 y = ex ; y(0) = 1
=⇒ Second order, linear, non-constant coeff’s
dy
3. x dx
− (1 + x)y = xy 2 + 5; y(0) = 1
=⇒ First order, non-linear, non-constant coeff’s
4. y 0 +tan(x)y =cos2 (x); y(0) = −1
=⇒ First order, linear, non-constant coeff’s
5. −y 00 + 2y 0 − 5y = 0; y(0) = 0, y 0 (0) = 1
=⇒ Second order, linear, constant coeff’s
6.
dy
dx
+ 3x2 y = x2 ; y(0) = 5
=⇒ First order, linear, non-constant coeff’s, separable
dy
7. y 1/2 dx
+ y 3/2 = 1; y(0) = 4
=⇒ First order, non-linear, constant coeff’s, separable, Bernoulli
8.
2
d2 y
dx2
dy
+ 4y = 0; y(0) = 1, y 0 (0) = 0
− 4 dx
=⇒ Second order, linear, constant coeff’s
Solution Techniques
For equations (4)-(8) above, (a) if the equation is first order, state all possible solution techniques
(give reasons), (b) solve each of the equations, regardless of order, by an appropriate technique.
1.4 y 0 + tan(x)y = cos2 (x); y(0) = −1
a) This DE is linear, thus integrating factor method will work. It is not separable and does not
have constant coeff’s. Thus no other solutions techniques will work.
b) The general solution is: y = c1 cos(x) + sin(x) cos(x)
1.5 −y 00 + 2y 0 − 5y = 0; y(0) = 0, y 0 (0) = 1
a) We only know one solution technique for second order equations, so don’t worry about this yet.
b) The general solution is: y = et (C1 cos(2t) + C2 sin(2t))
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1.6
dy
dx
+ 3x2 y = x2 ; y(0) = 5
a) This DE is linear, thus integrating factor method will work. Moreover, it is separable. Thus,
separation of variables will work as well.
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b) The general solution is: y = Ae−x + 1/3
dy
1.7 y 1/2 dx
+ y 3/2 = 1; y(0) = 4
a) This DE is non-linear, thus integrating factor method will not work. It is separable and and
Bernoulli. So either the Bernoulli method or separation of variables will work.
b) The general solution is: y = (C1 e−3x/2 + 1)2/3
1.8
d2 y
dx2
dy
− 4 dx
+ 4y = 0; y(0) = 1, y 0 (0) = 0
a) We only know one solution technique for second order equations, so don’t worry about this yet.
b) The general solution is: y = C1 e2x + C2 xe2x
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Existence and Uniqueness
Determine a region of the xy-plane for which the given differential equations would have a unique
solution passing through a point (x0 , y0 ) in that region.
1.
dy
dx
= y 2/3
p
=⇒ f (x, y) = 3 y 2 , continuous ∀x, y ∈ R.
2
=⇒ fy = 3 √
3 y , continuous ∀x, y ∈ R, y 6= 0.
dy
2. x dx
=y
=⇒ f (x, y) = xy , continuous ∀x, y ∈ R, x 6= 0.
=⇒ fy = x1 , continuous ∀x, y ∈ R, x 6= 0.
3. (1 + y 3 )y 0 = x2
x2
=⇒ f (x, y) = 1+y
3 , continuous ∀x, y ∈ R, y 6= −1.
=⇒ fy =
−3x2 y 2
,
(1+y 3 )2
continuous ∀x, y ∈ R, y 6= −1.
4. (x − y)y 0 = y + x
=⇒ f (x, y) = x+y
x−y , continuous ∀x, y ∈ R, x 6= y.
2x
=⇒ fy = (x−y)2 , continuous ∀x, y ∈ R, x 6= y.
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Phase Lines, Fixed Points, etc.
Consider the following first order differential equation:
y 0 = y 3 − 15y 2 + 50y.
(1)
Perform the following tasks: (a) find all fixed points and classify their stability, (b) draw the phase
line, (c) next to the phase line, draw sample solution curves in each region.
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Solution
a) Fixed points
y ∗ = 0, 5, 10. Where y ∗ = 0 is unstable, y ∗ = 5 stable and y ∗ = 10 is unstable.
c) Solution Curves
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Heating/Cooling
A small metal bar, whose initial temperature was 20◦ C, is dropped into a large container of boiling
water. How long will it take the bar to reach 90◦ C if it is known that its temperature increases 2◦
in one second?
Solution
The problem setup should be:
T 0 = k(T − 100); T (0) = 20, T (1) = 22
Thus the general solution is,
T = 100 + cekt .
Where you should find that,
c = −80, k = ln(39/40).
Finally, when T = 90, t = 82.1.
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6.1
“Challenge” Questions
Write your own tank problem.
The general solution to the differential equation is:
y(t) =
6.2
200 + t2 + C
100 + t
Bifurcation
We begin with the differential equation:
dP
= P (4 − P ) − h
dt
Thus, the fixed points are given by:
P∗ = 2 ±
√
4−h
Hence, there is only one fixed point (i.e., a bifurcation occurs) when h = 4.
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One More Thing
There are a few topics that are not included on this review, but that doesn’t mean they won’t be
on the exam. You are still responsible for all material covered in lecture!
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