MTH 352
Exam 1
Prof. Townsend
October 22, 2009
Your Name _________________________________________
Your Partner's Name _________________________________________
1) (30 points: 6 points each – 3 for linear, 3 for separable) Identify whether the following differential
equations are separable, linear, neither or both. To receive full credit you must put the differential equation
in the standard form for separable and/or linear differential equations, when appropriate. If separable, please
identify A(x) and B(x). If linear, please identify p(x) and q(x). If not linear or not separable, tell me how
you know.
Do not solve the equation; just put it in the appropriate form(s).
ID
a)
Exam A
Exam B
1 dy
+ x 2 csc (x ) = 2 xy cot (x ) multiply by x
x dx
dy
+ !2x 2 cot x y = !x 3 csc x
dx
linear, not separable
dy
y2
x
=
dx 1 + x
nonlinear, separable
!# 1 %#
!1%
"
& dx ( " 2 & dy
$y '
$# x 1 + x '#
2
dy 2x
y
!
+ 3
dx
y
x
{
nonlinear in y, not separable
b)
1 dy
= y 2 (1 + x )
x dx
nonlinear, separable
"1%
x 1 + x dx ! # 2 & dy = 0
$y '
{(
c)
d)
)}
dy
y
+ 3y =
both
dx
x
dy " 3
1%
+# 2 ! 3&y =0
dx $ x
x '
"3
"1%
1%
# 2 ! 3 & dx ! # & dy = 0
x '
$x
$y'
5 y ln (x )
dy
! y tan (x ) !
both
dx
x
5ln x &$
dy "$
+ #! tan x +
'y =0
dx $
x $
%
(
()
( ) $( dx = 0
!(
5ln x
!1$
dy
+
" %
"' tan x +
x
#y&
#(
()
x2
{
)
dy 3 y y 2
=
+
dx x 2 x3
nonlinear
not separable
x2
()
e)
(
( )}
dy
! x = xy divide by x 2
dx
%
&(
dy
! 2 y = x 2 y 2 divide by y
dx
dy
x ! 2 = x2 y
dx
"2 %
dy
+ !x y = # &
dx
$x'
linear, not separable
yx
{ }
x2
dy
= xy + x
dx
( )}
MTH 352
Exam 1
Prof. Townsend
dy 1 1
!
= y
dx x 2 x
"1%
dy " 1 %
+ #! & y = # 2 &
dx $ x '
$x '
linear, not separable
Note – you can spot both if q(x) for linear is 0.
October 22, 2009
dy
! x = xy
dx
see left = same problem
x2
MTH 352
Exam 1
Prof. Townsend
October 22, 2009
2 & 3) For problems 2 and 3 (20 points each), choose from the following list of eight differential equations.
One equation must be solved using separable techniques and another, by using the formalism for solving
linear differential equations. You may solve the same equation by two different methods, if appropriate.
You may not solve the same problem using the same method as your partner. Don't forget to use your
calculator or internet integrator (integrals.com) to perform integrations. Feel free to use one of the PCs in the
room.
a)
dy
+ x = x ( y + 1)
dx
dy 1 + x 2
= 2
dx
y
nonlinear
y 2 dy = 1 + x 2 dx
e)
dy
+ x = xy + x
dx
dy
= xy both
dx
Separable
dy
= xdx
y
()
{
3
y
x
= x+ +C
3
3
()
3
y x = 3x + x 3 + C '
( )
A x =x
}
3
B y =!
1
y
x2
ln | y |=
+C
2
y = C 'e
Linear
x2
2
dy
! xy = 0
dx
p x = !x
()
x2
2
()
!
()
x2
2
I x =e
y x =e
b) x
()
q x =0
{C }
dy
! 2 x 2 y = x 2 divide by x
dx
dy
! 2xy = x
dx
not separable
p x = !2x
()
()
q x =x
&
2 #
2
1
y x = e x $C + " e! x 2x dx '
2
%
(
2 "
2
1 2%
1
y x = e x #C ! e! x & = Ce x !
2
2
$
'
()
()
dy
= 3 x 2e ! y
dx
nonlinear
e y dy = 3x 2 dx
f)
e y = x3 + C
y x = ln x 3 + C
()
{
}
MTH 352
Exam 1
Prof. Townsend
dy 3
4
! y=
dx x
x
not separable
c)
()
!3
"
1 1 4 %
y x = x 3 #C + ! 3 dx &
2 x x '
$
()
()
dy
= 2 x (y 2 + 1)
dx
nonlinear
1
dy = 2xdx
2
y +1
g)
I x = e!3ln x = eln x = x !3 =
{
y x = x C + 2 " x dx
3
October 22, 2009
!4
}
1
x3
tan !1 y = x 2 + C
()
(
y x = tan x 2 + C
)
"$
x !3 &$
y x = x 3 #C + 2
'
!3 $(
$%
2
y x = Cx 3 !
3
()
()
dy
! x+ y
= 2e ( )
dx
nonlinear
e y dy = 2e! x dx
d)
e y = !2e! x + C
y x = ln C ! 2e! x
()
(
dy
= x3 ! 2 xy
dx
not separable
dy
+ 2xy = x 3
dx
h)
)
()
{
y x = e! x C + " e x x 3 dx
2
2
}
%
2 "
1 2
y x = e ! x #C + e x x 2 ! 1 &
2
$
'
2
1
y x = Ce! x + x 2 ! 1
2
(
()
(
()
)
)
4) (10 points) Evaluate C for one of your solutions in problems 2 and 3 using the condition that y(1)=1.
Then write the full solution where C has been replaced by its value.
5) (10 points) Solve one of the following problems. Do not choose that same one as your partner.
Example) You are given the following resisters and capacitors: …………..
In making a simple RC series circuits, which components would you choose to produce the minimum
response time? Justify your answer.
6) (10 points) Determine whether or not the given "solution" is a solution of the given differential
equation. Be sure to show any intermediate steps. You may perform the algebra steps on your TI but you
need to write down what you are doing.
Solution?
D.E.
MTH 352
Exam 1
Prof. Townsend
Exam A
1
x
yes - take derivative of y then plug
into LHS of D.E. You get the RHS
Exam B
y = xe! x
yes – see above
y = x2 +
October 22, 2009
x2
2
ex
2
dy
+ 1 = 2x 3
dx
dy
+ 2x 2 = 1
dx