Math 3301
Homework Set 1 – Solutions
10 Points
1. (2 pts) From the differential equation we can see
that the derivative will be zero at : y = −1,3,5 .
A sketch of the direction field and a few solutions is
shown to the right. From this we can see that the
long term behaviors are,
y ( 0) > 3
y (t ) → 5
−1 < y ( 0 ) ≤ 3
y (t ) → 3
−1
−1
y ( 0) =
y (t ) =
y ( 0 ) < −1
y ( t ) → −∞
3. (2 pts)
2 7x
4x
2
1
y
x
e
=+
(
)
x2 + 1
′
−2
⌠ 2
7x
 ( x + 1) y dx =∫ e dx
⌡
µ ( x)
y′ −
)
(
y ( x) =
−3 =y ( 0 ) =17 + c
6. (3 pts)
µ (t ) =
e
− 13 t
→
∫ (e
1
7
(x
→
2
2
1t
=+
( x2 2)
−2
2
y ( x ) =17 ( x 2 + 1) e7 x − 227 ( x 2 + 1)
2
⇒
′
− 1t
y dt =
∫ e 3 ( 43 t − 53 ) dt →
y (t ) =
− ( 4t + 7 ) + ce 3
)
−2
2
)
(
−2ln x 2 + 2
e
=
+ 1) y =17 e7 x + c
+ 1) e7 x + c ( x 2 + 1)
c =− 227
− 13 t
(x
4x
⌠
 − 2 dx
⌡ x +1
e
=
e
− 13 t
y=
−e
− 13 t
2
( 4t + 7 ) + c
y (t ) =
− ( 4t + 7 ) + ( y0 + 7 ) e 3
1t
⇒
Okay, in order for this situation to happen the function will have a value of zero at this point and
because the solution is clearly continuous and will have a continuous derivative it must also occur at a
relative extrema (don’t know if it will be a maximum or a minimum, but that’s not important) and so we
also know that at this point the derivative must be zero. So, let’s call the point te and we know that
′ ( te ) 0 . So, all we need to do is plug into the differential equation and solve for te . Doing
y=
( te ) y=
this gives,
3 y ′ ( te ) − y ( te ) =
4te − 5
0=
4te − 5
5
te =
4
Now we know that y ( 54 ) = 0 so plug this into the solution we found above and solve for y0 .
0=
y ( 54 ) =
−12 + ( y0 + 7 ) e 12
5
⇒
y0 =
0.9108875624
Math 3301
Homework Set 1 – Solutions
10 Points
7. (3 pts)
µ (t ) =
e −3t
−t
− 4t
⌠ ( e −3t y )′ dt =
∫ 4e − 7e dt ⇒
⌡
−4e − t + 74 e − 4t + c
e −3t y =
y ( t ) =74 e − t − 4e 2 t + ( 414 + α − 2α 2 ) e3t
y (t ) =
−4e 2 t + 74 e − t + ce3t
Okay, the first time will always go to zero, the second term will always go to minus infinity. The last
term will vary depending the sign of the coefficient. Note however that if the third term isn’t there (i.e.
the coefficient is zero) then the second term will determine long term behavior. If the third term goes
to positive infinity (i.e. the coefficient is positive) then the solution will go to positive infinity (the
exponent on the third exponential is larger than the exponent on the second exponential). Finally, if the
third term goes to negative infinity (i.e. the coefficient is negative) then because this exponent is larger
than the exponent of the second term it will dominate and the solution will go to negative infinity.
So, all we need to do is set the coefficient of the third term equal to zero and determine where it is
positive/negative. Here’s a table giving all these results.
1− 83
4
< α < 1+ 483
−2.0276 < α < 2.5276
y (t ) → ∞
α ≥ 1+ 483 , α ≤ 1− 483
y ( t ) → −∞
y (t ) → ∞
α ≥ 2.5276, α ≤ −2.0276
y ( t ) → −∞
Not Graded
2. From the differential equation we can see that
the derivative will be zero at : y = −2, 0, 2 .
A sketch of the direction field and a few solutions is
shown to the right. From this we can see that the
long term behaviors are,
y ( 0) > 0
y (t ) → 2
−2 < y ( 0 ) ≤ 0
y (t ) → 0
−2
−2
y ( 0) =
y (t ) =
y ( 0 ) < −2
y ( t ) → −∞
4.
y′ + ( 4 + 6t ) y=
⌠ t 6e 4 t y )′ =
dt
⌡(
2e −t 10
− 6
t5
t
∫ 2te
3t
− 10e 4t dt
µ ( t )= e ∫
→
4 + 6t dt
= e
4 t + 6 ln t
= t 6e 4 t
t
t 6e 4=
y 2e3t ( 3t − 19 ) − 52 e 4t + c
=
y ( t ) 2t −6e −t ( 3t − 19 ) − 52 t −6 + ct −6e −4t
Math 3301
Homework Set 1 – Solutions
−2= y (1=
) 2e−1 ( 92 ) − 52 + ce−4 →
c=
10 Points
1
2
e 4 − 94 e3
=
y ( t ) 2t −6e −t ( 3t − 91 ) − 52 t −6 + ( 12 e 4 − 94 e3 ) t −6e −4t
5. We know from Calc I that relative extrema occur at critical points and critical points are those points
where the derivative is zero or doesn’t exist. However, we’ve been told that the derivative exists and is
continuous everywhere so that means that at the critical point that gives the relative maximum, let’s call
it tc = 0.8 , we must have y′ ( 0.8 ) = 0 and we want to determine y ( 0.8 ) so all we need to do is
“plug” tc = 0.8 into the differential equation use what we know about the derivative and solve for
y ( 0.8 ) .
y′ ( 0.8 ) + 32 y ( 0.8 ) =
e −2( 0.8) − e0.8
⇒
(
)
−2( 0.8 )
2
y ( 0.8 ) =
− e0.8 =
−1.3491
3 e