# M427K (55590) Midterm #1 Solutions

## Transcription

M427K (55590) Midterm #1 Solutions
```M427K (55590)
Midterm #1
Solutions
1. Consider the linear ODE
t
dy
+ y = t2 .
dt
(a) Suppose you are given the initial condition y(t0 ) = y0 . Find all values of t0 that guarantee that
the initial value problem has a unique solution.
Solution.
First put the ODE into standard form:
dy 1
+ y = t.
dt
t
We can identify p(t) = 1t , which is continuous everywhere except at t = 0, and g(t) = t, which
is continuous everywhere.
Thus choosing t0 to be any non-zero value will guarantee that the IVP has a unique solution.
(b) Suppose t0 = 2. Find the largest interval on which the initial value problem is guaranteed to
have a unique solution.
Solution.
Since t0 = 2 > 0, the IVP will have a unique solution on the interval (0, ∞).
(c) Find the general solution of the ODE.
Solution.
The equation needs to be in standard form, as in part (a). This is a linear first-order ODE, so
we look for an integrating factor.
Z
µ(t) = exp
1
dt = eln t = t.
t
Multiply by µ(t):
d
(ty) = t2 .
dt
Integrate:
ty =
1 3
t + C.
3
Solve for y:
y(t) =
1 2 C
t + .
3
t
2. Consider the ODE
dy
= −y 2 .
dt
(a) Sketch the right-hand side f (y) = −y 2 .
Solutions.
1
(b) Draw the phase line.
Solutions.
(c) Find all equilibrium points and classify each one as stable, unstable, or semistable.
The function f (y) vanishes at y = 0, so y = 0 is an equilibrium. From the phase plot, it is a
semistable equilibrium.
Solutions.
(d) Sketch several solutions of the ODE that exhibit the different types of behaviour that can be
expected.
2
3. Consider the ODE
−t − y 2 +
dy
= 0.
dt
(a) Is the equation linear or nonlinear?
Solution.
The equation is nonlinear since it contains the term y 2 .
(b) Is the equation separable?
Solution.
The equation is not separable since it involves a sum of a function of t and a function of y.
Solution.
To check if the equation is exact, identify M (t, y) = −t − y 2 and N (t, y) = 1. Then compute
My = −2y,
Nt = 0.
These are not equal so the equation is not exact.
(d) Given the initial condition y(0) = 1, use Euler’s method with h = 0.1 to approximate y(0.1).
Solution.
The equation can be rewritten as
dy
= t + y2
dt
so that f (t, y) = t + y 2 . We also know that t0 = 0 and y0 = 1.
Euler’s method involves computing
yn+1 = yn + h(tn + yn2 ).
Thus
y(0.1) ≈ y1 = 1 + 0.1(0 + 12 ) = 1.1.
4. Solve the following initial value problem using any of the techniques learned in this course.
y 00 − 3y 0 + 2y = 6e−t ,
y(0) = 1,
y 0 (0) = 2
Solution.
The characteristic equation for this problem is
r2 − 3r + 2 = (r − 1)(r − 2) = 0,
which has the two real roots r = 1, 2. Thus the homogeneous solution is
yh (t) = C1 et + C2 e2t .
Now we look for a particular solution of the form
yp (t) = Ae−t ,
yp0 (t) = −Ae−t ,
3
yp00 (t) = Ae−t .
Substitute this into the ODE:
yp00 (t) − 3yp0 (t) + 2yp (t) = (A + 3A + 2A)e−t = 6Ae−t .
Setting this equal to the given right-hand side 6e−t , we obtain A = 1. Thus the general solution is
y(t) = yh (t) + yp (t) = C1 et + C2 e2t + e−t .
We we also need the derivative of this:
y 0 (t) = C1 et + 2C2 e2t − e−t .
The initial conditions require
y(0) = C1 + C2 + 1 = 1
⇒
C2 = −C1 ,
0
y (0) = C1 + 2C2 − 1 = 2 ⇒ −C1 = 3,
which yields C1 = −3 and C2 = 3. Thus the solution of the IVP is
y(t) = −3et + 3e2t + e−t .
4
```