Math 215
Spring 2010
Assignment 2
You are highly recommended to own a textbook, Boyce-DiPrima or another, to see a comprehensive treatment of the subjects and more examples.
§2.6: 7, 13, 25, 28, §2.7: 2(a), 4(a), §3.1: 6, 12, 21, 26(a)
• (§2.6 #7) Determine if the following equation is exact. If it is exact, find the solution.
(ex sin y − 2y sin x)dx + (ex cos y + 2 cos x)dy = 0.
Answer.
For M = ex sin y − 2y sin x and N = ex cos y + 2 cos x, we have My =
ex cos y − 2 sin x = Nx and thus the DE is exact. Let
Z
Z
ψ = M (x, y)dx = (ex sin y − 2y sin x)dx = ex sin y + 2y cos x + h(y).
ψy = ex cos y + 2 cos x + h0 (y) = ex cos y + 2 cos x.
Thus h0 (y) = 0 and h is constant. Hence an implicit solution of the DE is
ψ(x, y) = ex sin y + 2y cos x = c.
• (§2.6 #13) Solve the given initial value problem and determine at least approximately
where the solution is valid.
(2x − y)dx + (2y − x)dy = 0,
y(1) = 3.
Answer.
M = 2x − y and N = 2y − x. Since My = −1 = Nx , the equation is
exact. Integrating M with respect to x while holding y constant yields ψ(x, y) =
x2 − xy + h(y). Now ψy = 0 − x + h0 (y) = N = 2y − x. thus h0 (y) = 2y and h(y) = y 2 .
Thus the solution is given implicitly as ψ(x, y) = x2 − xy + y 2 = c. Invoking the initial
condition y(1) = 3, the specific solution is
x2 − xy + y 2 = 7.
√
The explicit form of the solution is y(x)
=
(x
+
28 − 3x2 )/2. Hence the solution is
q
valid as long as 3x2 ≤ 28, i.e., |x| ≤ 28
3 .
• (§2.6 #25) Find an integrating factor for the following equation and solve it.
(3x2 y + 2xy + y 3 )dx + (x2 + y 2 )dy = 0.
Answer.
The equation is not exact since My = 3x2 + 2x + 3y 2 6= Nx = 2x, so we
must attempt to find an integrating factor. Try µ = µ(x), we get (the new)
M = (3x2 y + 2xy + y 3 )µ(x),
N = (x2 + y 2 )µ(x),
My = (3x2 + 2x + 3y 2 )µ = Nx = 2xµ + (x2 + y 2 )µ0 ,
3µ = µ0 ,
which has a solution µ = e3x . Thus
Z
Z
1
ψ = M dx = (3x2 y + 2xy + y 3 )e3x dx = (x2 y + y 3 )e3x + h(y),
3
ψy = (x2 + y 2 )e3x + h0 (y) = (x2 + y 2 )e3x ,
thus h0 (y) = 0, Taking h = 0, the solution is given implicitly as
1
ψ(x, y) = (x2 y + y 3 )e3x = c.
3
• (§2.6 #28) Find an integrating factor for the following equation and solve it.
y dx + (2xy − e−2y )dy = 0.
Answer.
The equation is not exact since My = 1 6= Nx = 2y.
Try µ = µ(x), then
(yµ)y = µ = [(2xy − e−2y )µ]x = 2yµ + (2xy − e−2y )µ0 ,
which is not an ODE for µ(x). Try µ = µ(y), then
(yµ)y = µ + yµ0 = [(2xy − e−2y )µ]x = 2yµ,
resulting in
1
µ0 + ( − 2)µ = 0,
y
which is a linear separable ODE with a solution
Z
1
µ(y) = exp −( − 2)dy = exp(− ln y + 2y) = e2y /y.
y
If y 6= 0, the new equation is e2y dx + (2xe2y − 1/y)dy = 0. Thus
Z
ψ = e2y dx = xe2y + h(y),
2
ψy = 2xe2y + h0 (y) = N = 2xe2y − 1/y,
thus h0 (y) = −1/y and h(y) = − ln |y|. we get
2xe2y − ln |y| = c.
If y(x0 ) = 0 for some x0 , then y(x) = 0 for all x is the solution.
• (§2.7 #2(a)) Find approximate values of the solution of the following initial value
problem at t = 0.1, 0.2, 0.3, and 0.4 using the Euler method with h = 0.1.
y 0 = 2y − 1,
Answer.
y(0) = 1.
We have the recursive formula y0 = 1 and yk+1 = yk + 0.1(2yk − 1). Thus
y1 = 1.1,
y2 = 1.22,
y3 = 1.364,
y4 = 1.5368.
• (§2.7 #4(a)) Find approximate values of the solution of the following initial value
problem at t = 0.1, 0.2, 0.3, and 0.4 using the Euler method with h = 0.1.
y 0 = 3 cos t − 2y,
y(0) = 0.
Answer.
We have the recursive formula tk = 0.1k, y0 = 0 and yk+1 = yk +
0.1(3 cos tk − 2yk ). Thus
y1 = 0.3,
y2 = 0.538501,
y3 = 0.724821,
y4 = 0.866458.
• (§3.1 #6) Find the general solution of
4y 00 − 9y = 0.
Answer. The characteristic equation is 4r2 − 9 = 0, with roots r = ±3/2. Therefore
the general solution is y(t) = c1 e−3t/2 + c2 e3t/2 .
• (§3.1 #12) Find the the solution of the following initial value problem. Sketch the
graph of the solution and describe its behavior as t increases.
y 00 + 3y 0 = 0,
y(0) = −2,
y 0 (0) = 3.
Answer. The characteristic equation is r2 + 3r = 0, with roots r = −3, 0. Therefore
the general solution is y(t) = c1 + c2 e−3t , with derivative y 0 (t) = −3c2 e−3t . The initial
conditions give −2 = c1 + c2 and 3 = −3c2 . Thus c2 = −1, c1 = −1, and the solution
is y(t) = −1 − e−3t .
The graph is a monotone increasing function which converges to y = −1 tangentially
from below as t → ∞, and to −∞ as t → −∞.
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• (§3.1 #21) Solve the initial value problem y 00 − y 0 − 2y = 0, y(0) = α, y 0 (0) = 2. Then
find α so that the solution approaches zero as t → ∞.
Answer.
The characteristic equation is r2 − r − 2 = 0, with roots r = −1, 2. The
general solution is y = c1 e−t + c2 e2t . The initial conditions give α = c1 + c2 and
2 = −c1 + 2c2 , so adding them we find 3c2 = 2 + α. If y(t) is to approach zero as
t → ∞, c2 must be zero. Thus α = −2.
• (§3.1 #26(a)) Solve the initial value problem for β > 0:
y 00 + 5y 0 + 6y = 0,
y(0) = 2,
y 0 (0) = β.
Answer. The characteristic equation is r2 + 5r + 6 = 0, with roots r = −2, −3. The
general solution is y = c1 e−2t + c2 e−3t . The initial conditions give 2 = c1 + c2 and
β = −2c1 − 3c2 , so c1 = 6 + β, c2 = −4 − β, and y = (6 + β)e−2t + (−4 − β)e−3t .
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