Solved Problems on Elementary Differential Equations - Assignment 5 | Math 3, Assignments of Mathematics

Material Type: Assignment; Class: ELEM DIFF EQUATIONS; Subject: Mathematics; University: University of California - Irvine; Term: Winter 2006;

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MATH 3D Homework 5 Name: Solutions
Feb. 10, 2006
hw5.pdf Problems
2. Find the general solution of
1 + µt
ysin(y)dy
dt = 0.
Solution: (10 points) This is a first-order, nonlinear problem, and it doesn’t
look separable, so we look to the method of exact equations. However, if M= 1
and
N=t
ysin(y),
then My6=Nt, and the equation is not exact. So, we multiply by µand see if
we can make it exact:
µ+µµt
ysin yy0= 0.
Then we require that
µy=µµµt
ysin y¶¶t
=µtµt
ysin y+1
yµ.
Suppose that µ=µ(y), and so µt= 0 and µy=µ0. Then
µ01
yµ= 0 =µ=eR1
ydy =eln|y|=|y|.
So, the modified equation is
|y|+µt|y|
y |y|sin yy0= 0.
Suppose first that y > 0. Then the ODE is
y+ (tysin y)y0= 0.
If M=yand N=tysin y, then we require that
ϕt=M=y
ϕy=N=tysin y.
By the first equation,
ϕ=Zϕtdt =ty +f(y) =ϕy=t+f0(y) = tysin y.
So,
f0(y) = ysin y=f(y) = ycos ysin y+c
So,
ϕ(t, y) = ty +ycos ysin y+c= 0
gives the general solution.
Document URL: http://math.uci.edu/~pmacklin/Math3Dwinter2006.html
Date: February 22, 2006 Page 1 of 2
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MATH 3D Homework 5 Name: Solutions

Feb. 10, 2006

hw5.pdf Problems

  1. Find the general solution of

1 +

t y

− sin(y)

dy dt

Solution: (10 points) This is a first-order, nonlinear problem, and it doesn’t look separable, so we look to the method of exact equations. However, if M = 1 and N =

t y

− sin(y),

then My 6 = Nt, and the equation is not exact. So, we multiply by μ and see if we can make it exact: μ + μ

t y

− sin y

y′^ = 0.

Then we require that

μy =

μ

t y

− sin y

t = μt

t y

− sin y

y

μ.

Suppose that μ = μ(y), and so μt = 0 and μy = μ′. Then

μ′^ −

y

μ = 0 =⇒ μ = e

∫ (^1) y dy^ = eln|y|^ = |y|.

So, the modified equation is

|y| +

t |y| y

− |y| sin y

y′^ = 0.

Suppose first that y > 0. Then the ODE is y + (t − y sin y) y′^ = 0.

If M = y and N = t − y sin y, then we require that ϕt = M = y ϕy = N = t − y sin y.

By the first equation,

ϕ =

ϕt dt = ty + f (y) =⇒ ϕy = t + f ′(y) = t − y sin y.

So, f ′(y) = −y sin y =⇒ f (y) = y cos y − sin y + c So, ϕ(t, y) = ty + y cos y − sin y + c = 0 gives the general solution.

Document URL: http://math.uci.edu/~pmacklin/Math3Dwinter2006.html Date: February 22, 2006 Page 1 of 2

Section 2.1 Problems

  1. Let y 1 and y 2 be solutions of Bessel’s equation

t^2 y′′^ + ty′^ + (t^2 − n^2 )y = 0

on the interval 0 < t < ∞, with y 1 (1) = 1, y′ 1 (1) = 0, y 2 (1) = 0, and y′ 2 (1) = 1. Compute W y 1 , y 2 .

Solution: (10 points) Recall that if y 1 and y 2 are solutions to

y′′^ + py′^ + qy = 0,

on (a, b), then the Wronskian W (t) = W y 1 , y 2 solves W ′^ + pW = 0 on (a, b). Our differential equation is not in the proper form; we have to first divide by t^2 :

y′′^ +

t

y +

t^2 n^2

y = 0.

Then W solves W ′^ + (^1) t W = 0 on (0, ∞), i.e.,

W (t) = W (1)e−^

∫ (^) t 11 s ds^ = W (1)eln 1−ln^ t^ = W (1)^1 t

Now, we need W (1):

W (1) = det

y 1 (1) y 2 (1) y 1 ′(1) y′ 2 (1)

= y 1 (1)y′ 2 (1) − y 2 (1)y 1 ′(1) = 1.

Thus, W (t) =

t

Completeness Points

(5 points) One point per seriously-attempted, assigned problem.

Document URL: http://math.uci.edu/~pmacklin/Math3Dwinter2006.html Date: February 22, 2006 Page 2 of 2