1. Solve the following initial value problem:
dy
x
= −3y +
dx
y
y(0) = −5 .
——————————————————————————————–
Bernoulli.
dy
+ 3y = x · y −1 .
dx
The change of variables y = v 1/2 leads to:
dv
+ 2 · 3v = 2x .
dx
The simplest integrating factor here is e6x , and after multiplying and
then simplifying the left hand side (using the product rule in reverse),
you should get to:
(e6x v)′ = 2xe6x .
Then taking antiderivatives (including doing your integration by parts
correctly) gets you to here:
∫
x
1
6x
e v = 2xe6x dx = e6x − e6x + C .
3
18
Now using y 2 = v and doing a bit of algebra gives us:
y2 =
x
1
−
+ Ce−6x .
3 18
The initial data can be used here to find C :
25 = 0 −
1
+C
18
and so C = 25 +
1
1
451
= 25 =
.
18
18
18
Taking the negative square root because of the initial data leads to:
√
√
6x − 1 + 451e−6x
1 6x − 1 + 451e−6x
y=−
=−
.
18
3
2
2. Solve the following initial value problem:
dy
3t2
=
dt
2y − 1
y(0) = −1 .
——————————————————————————————–
Separable.
∫
∫
(2y − 1) dy = (3t2 ) dt ,
so
y 2 − y = t3 + C.
Plugging in initial conditions leads to C = 2. To solve for y, write it in
the form:
y 2 − y + (−t3 − 2) = 0
so that the quadratic formula can be used. That, along with another
use of the initial data leads to:
√
√
−(−1) − 1 − 4(−t3 − 2)
1 − 9 + 4t3
y=
=
.
2
2
3. Solve the following initial value problem for t > 0 :
dy
y
= − + 4 sin(2t)
dt
t
y
(π )
2
=1.
——————————————————————————————–
Linear.
dy 1
+ y = 4 sin(2t).
dt
t
Finding the integrating factor:
∫
µ(t) = e
1
t
dt
= eln t = t.
(Since t > 0, we can ignore absolute values.) Multiplying through gives
us:
dy
t + y = 4t sin(2t) ,
dt
which is the same as:
(ty)′ = 4t sin(2t) .
∫
So:
ty =
4t sin(2t) dt .
After integrating by parts we get:
ty = sin(2t) − 2t cos(2t) + C .
Plugging in the initial data we get:
π
= sin(π) − π cos(π) + C = 0 + π + C ,
2
and so C = − π2 . Thus:
sin(2t) − 2t cos(2t) −
y(t) =
t
π
2
=
sin(2t)
π
− 2 cos(2t) −
.
t
2t
4. Find the general solution of the following ODE:
3x
dy
=
.
dx
y − 2x
You may leave your final answer in implicit form.
——————————————————————————————–
Homogeneous.
Use the change of variables y = vx to get to:
x
dv
3
+v =
.
dx
v−2
Then
dv
3
v(v − 2)
−v 2 + 2v + 3
=
−
=
.
dx
v−2
v−2
v−2
A little bit of algebra will then lead to:
∫
∫
2−v
dx
dv =
.
(v − 3)(v + 1)
x
x
The partial fractions decomposition for the integrand in the left hand
side is:
(
)
(
)
3
1
1
1
−
−
4 v+1
4 v−3
so integrating both sides gives us:
3
1
− ln |v + 1| − ln |v − 3| = ln |x| + C .
4
4
After bringing the ln |x| to the left, using v = y/x, multiplying by −4
and absorbing a −4 into the constant C, we get to:
3 ln |(y/x) + 1| + ln |(y/x) − 3| + 4 ln |x| = C .
By standard identities for the natural logarithm, this can be simplified
nicely to get:
(
)
ln |y + x|3 · |y − 3x| = C .
This is already very, very nice, but if you like, you could exponentiate
it and simplify to get:
(y + x)3 · (y − 3x) = C ,
which even includes the solution y = −x.
5. Solve the following initial value problem:
dy
6x2 − y
=
dx
x + 3y 2
y(2) = 1 .
You may leave your final answer in implicit form.
——————————————————————————————–
Exact.
dy
(y − 6x2 ) + (x + 3y 2 )
=0
dx
has the form
dy
=0,
M (x, y) + N (x, y)
dx
with My = 1 = Nx , which verifies that this equation is exact.
Then
∫
∫
F (x, y) = M (x, y) dx = (y − 6x2 ) dx = xy − 2x3 + C̃(y) ,
and
∫
F (x, y) =
∫
N (x, y) dy =
(x + 3y 2 ) dy = xy + y 3 + C̄(x) .
Matching up terms appropriately gives us:
F (x, y) = y 3 + xy − 2x3 = C .
The initial data give us:
1 + 2 − 16 = C
so
C = −13,
and thus:
y 3 + xy − 2x3 = −13 or y 3 + xy − 2x3 + 13 = 0
is the final answer.
6. For the following initial value problem:
dy
= cos(t3 y 4 )
dt
y(1) = −3
write down the standard Euler iteration with a stepsize of 0.01 (Explicitly give y0 and give yn+1 as a function of yn and n.) In terms of this
scheme, what would you use to approximate y(2.5)?
——————————————————————————————–
y0 = −3.
t0 = 1.
Since tn+1 = tn + .01, we have: tn = 1 + .01n. Then:
yn+1 = yn + .01 cos(t3n yn4 )
Or:
yn+1 = yn + .01 cos((1 + .01n)3 yn4 )
If 2.5 = tn = 1 + .01n, then n = 150. Thus,
y(2.5) ≈ y150 .
7. Compute and/or simplify the following. (Your answer must have the
form a + bi where a and b are real numbers or real functions.)
(a) (3 + 4i) + (5 − 6i) = 8 − 2i.
(b)
2 − 3i
=
4 + 5i
(
2 − 3i
4 + 5i
) (
)
4 − 5i
−7 − 22i
7
22
·
=
=− − i.
4 − 5i
16 + 25
41 41
(c)
e2x+3iy = e2x e3iy = e2x (cos(3y)+i sin(3y)) = e2x cos(3y)+ie2x sin(3y) .
8. (a) Find the equilibrium solutions to the problem:
dy
= y 2 − 7y + 10 .
dt
Determine whether each equilibrium solution is stable, unstable,
or semistable.
————————————————————————————
y 2 − 7y + 10 = (y − 5)(y − 2)
so y ≡ 2 and y ≡ 5 are equilibrium solutions.
Since y is increasing for −∞ < y < 2 and 5 < y < ∞ and since y
is decreasing for 2 < y < 5, we have a stable equilibrium at y ≡ 2
and an unstable equilibrium at y ≡ 5.
————————————————————————————
(b) The equation:
dy
2
6
= (5 − y)ey (sin y) log(3 + y 4 )
dt
has the equilibrium solution y ≡ 5. Is there a solution to that
differential equation with initial data y(0) = 1? If no, then why
not? If yes, then for the solution to that initial value problem is
there a value T such that y(T ) = 7? Justify your answers! If you
quote any theorems then state the hypotheses of the theorem that
you need to check.
——————————————————————————————–
Although the right hand side is a mess, it is certainly continuous and
differentiable, and therefore, the existence-uniqueness theorem applies.
Thus, there exists a solution the initial value problem with y(0) = 1.
Because of the uniqueness part of the theorem, the solution cannot
cross the equilibrium solution y ≡ 5. Therefore, since it starts below
that solution, it will stay below that solution forever. (So in particular,
it will never equal 7.)