MATH 152
L1 & L2
Applied Linear Algebra and Differential Equations
Spring 200405
¥
Week 02 Worksheet Solutions
:
Firstorder differential equations
Q1.
(Change of variables)
Consider the firstorder differential equation
1 +
x
y

sin
y
¶
dy
dx
= 0
,
which is neither linear nor separable. The purpose of this problem is to show you a trick.
(a) Instead of thinking of the variable
y
as a function of
x
, suppose we think of the variable
x
as a
function of
y
. The derivative of
x
with respect to
y
is
dx
dy
, the reciprocal of
dy
dx
. Rewrite the given
equation so that it is an equation for
dx
dy
.
Solution
x
y

sin
y
¶
=

dx
dy
,
dx
dy
+
x
y
=
sin
y.
(1)
(b) Is your rewritten equation linear, separable or both?
Solution
The equation (1) is a firstorder linear ordinary differential equation for
x
(
y
). However,
it is a nonseparable equation.
(c) Solve your differential equation for
x
as a function of
y
. This gives you an implicit function for
y
in terms of
x
.
Solution
Consider the equation (1) for
x
(
y
). The integrating factor is
exp
•Z
1
y
dy
‚
=
e
ln
y
=
y.
Then consider
d
dy
(
yx
) =
y
dx
dy
+
x
y
¶
=
y
sin
y.
Integrating with respect to
y
gives
yx
=
Z
y
sin
y dy
=

Z
y d
(cos
y
)
=

y
cos
y
+
Z
cos
y dy
=

y
cos
y
+ sin
y
+
C.
The implicit solution for
y
in terms of
x
is given by
xy
+
y
cos
y

sin
y
=
C
.
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Q2.
(Homogeneous equation)
In the previous problem, we showed how one sometimes converts nonlinear
differential equations into linear differential equations. This problem shows you another trick to convert
a certain type of differential equation into a separable equation. Let us consider the equation
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 Spring '10
 KCC
 Math, Linear Algebra, Algebra, Equations, Elementary algebra, Linear map, dy

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