M427K (54315) Midterm #1 Solutions

M427K (54315)
Midterm #1
Solutions
1. Find the general solution of the following first order ODEs.
−2t
(a) t dy
dt + 2ty = e
Solution.
Write the linear equation in standard form:
dy
1
+ 2y = e−2t .
dt
t
Identify p(t) = 2 and compute the integrating factor
R
µ(t) = e
2 dt
= e2t .
Multiply by the integrating factor to obtain
d 2t 1
ye = .
dt
t
Integrating:
ye2t = ln |t| + C
and the general solution is
y = e−2t (ln |t| + C).
(b)
dy
dt
= e−2y sin(3t)
Solution.
The equation is separable, and can be expressed as
Z
Z
2y
e dy = sin(3t) dt.
Integrating,
1 2y
1
e = − cos(3t) + C.
2
3
Solving for y:
y=
1
2
ln − cos(3t) + 2C .
2
3
2. Consider the ODE
dy
= y 2 − 1.
dt
(a) Sketch the right-hand side f (y) = y 2 − 1.
Solution.
(b) Draw the phase line.
Solution.
1
(c) Find all equilibrium points and classify each one as stable, unstable, or semistable.
Solution.
There is an unstable equilibrium at y = 1 and a stable equilibrium at y = −1.
(d) Sketch several solutions of the ODE that exhibit the different types of behaviour that can be
expected.
Solution.
3. Consider the ODE x2 y 00 − 5xy 0 + 9y = 0.
(a) Verify that y1 = x3 is a solution of this ODE.
Solution.
Compute the derivatives of y1 :
y10 = 3x2 ,
y100 = 6x.
Substitute these into the ODE:
x2 y100 − 5xy10 + 9y1 = x2 (6x) − 5x(3x2 ) + 9(x3 ) = 0.
Thus y1 is a solution.
(b) Use the method of reduction of order to find a second solution y2 of the ODE.
Solution.
Look for a solution of the form y2 = x3 v. The derivatives are
y20 = x3 v 0 + 3x2 v,
y200 = x3 v 00 + 6x2 v 0 + 6xv.
Substitute into the ODE:
x2 y200 − 5xy20 + 9y2 = x2 (x3 v 00 + 6x2 v 0 + 6xv) − 5x(x3 v 0 + 3x2 v) + 9x3 v
= x5 v 00 + x4 v 0 = 0.
Substitute w = v 0 :
x5 w0 + x4 w = 0.
This is separable so we can write
Z
Z
dx
dw
=−
.
w
x
Integrate this to obtain
ln w = − ln x,
Then we can obtain
Z
v=
⇒
w=
1
.
x
1
dx = ln |x| .
x
Thus a second solution is
y2 = y1 v = x3 ln |x| .
4. Solve the following initial value problem using any of the techniques learned in this class.
y 00 − 2y 0 + 2y = sin t,
y(0) = y 0 (0) = 1.
Solution.
The characteristic equation is
r2 − 2r + 2 = 0
which has the roots
r=
2±
√
4−8
= 1 ± i.
2
The homogeneous solution is
yh (t) = C1 et cos t + C2 et sin t.
Look for a particular solution of the form
yp (t) = A cos t + B sin t,
yp0 (t) = −A sin t + B cos t,
yp00 (t) = −A cos t − B sin t.
Substitute this into the ODE:
yp00 − 2yp0 + 2yp = (A − 2B) cos t + (2A + B) sin t.
Require this to equal sin t, which yields A − 2B = 0 and 2A + B = 1. These have the solution
A = 2/5, B = 1/5.
The general solution is
2
1
y(t) = yh (t) + yp (t) = C1 et cos t + C2 et sin t + cos t + sin t.
5
5
The derivative of this is
y 0 (t) = C1 et (cos t − sin t) + C2 et (sin t + cos t) −
2
1
sin t + cos t.
5
5
The IC require
2
1
= 1, y 0 (0) = C1 + C2 + = 1.
5
5
The solutions are C1 = 3/5 and C2 = 1/5. Thus the solution of the IVP is
1
2
1
3
y(t) = et cos t + et sin t + cos t + sin t.
5
5
5
5
y(0) = C1 +