Exam Basics
1
There will be 9 questions.
2
The first 3 are on pre-midterm material.
3
The next 1 is a mix of old and new material.
4
The last 5 questions will be on new material since the
midterm.
5
60 points total (+3 bonus points)
Katie Walsh
Math 20D Final Exam Review
First Order Differential Equations
1
Linear First Order Differential Equations
1
2
3
2
y 0 + p(t)y =Rg (t)
Let µ(t) =R e p(t)dt
y=1/µ(t) µ(t)g (t)dt
Separable First Order Differential Equations
1
2
M(x)
+ N(y )dyR/dx = 0
R
N(y )dy = − M(x)dx
Katie Walsh
Math 20D Final Exam Review
Exact and Integrating Factor for Exact
A differential equation of the form
M(x, y )dx + N(x, y )dy = 0
is exact if
M y = Nx
where My is the partial derivative of the function M(x, y ) with
respect to y and Nx is the partial derivative of N(x, y ) with
respect to y .
If the equation is exact then we can find some function f such that
fx = M and fy = N. Then the solution to the differential equation
is f = C .
Katie Walsh
Math 20D Final Exam Review
Sometimes an equation is not exact but we can use an integrating
factor to make it exact. There are two types of problems you
should be able to do.
1 Equations that can be made exact by multiplying by a function
µ(x) that only depends on x.
This is true only if
My − Nx
N
only depends on N. In this case
µ(x)0 =
My − Nx
µ(x)
N
e.g.
y 0 = e 2x + y − 1
Katie Walsh
Math 20D Final Exam Review
2 Equations that can be made exact using a given integrating
factor.
e.g.
x 2 y 3 + x(1 + y 2 )y 0 = 0
µ(x, y ) = 1/(xy 3 )
Katie Walsh
Math 20D Final Exam Review
Examples of first order differential equations
1
dy /dx =
2
dy /dx =
x 3 −2y
x
2x+y
3+3y 2 −x
3
dy /dx =
4x 3 +1
y (2+3y )
4
For more examples see Pgs 132-133
Katie Walsh
Math 20D Final Exam Review
Autonomous Equations
1
Graph dy /dt as a function of y . Focus on zeros and where
the function is positive and negative.
2
Graph the phase line (number line with equilibrium points
marked and draw arrows to represent when solutions are
increasing or decreasing)
3
Classify the stationary points as stable, unstable or
semi-stable.
4
Sketch several solutions based on the increasing and
decreasing information from the phase line.
Example: dy /dt = y 2 (y 2 − 1) where −∞ < y0 < ∞
Katie Walsh
Math 20D Final Exam Review
Figure: Plot of dy /dt as a function of y
2
unstable
1
semistable
0
stable
-1
-2
Figure: Phase Line and sketch of solutions
Katie Walsh
Math 20D Final Exam Review
Figure: Actual Solutions
Follow-up Question: What happens to y as t → ∞?
This depends on the initial condition! Let y (t0 ) = y0 .
For y0 > 1, y → ∞.
For 0 < y0 < 1, y → 0.
For y0 < 0, y → −1 .
Katie Walsh
Math 20D Final Exam Review
Theorems about guaranteed solution intervals
First Order
Theorem
If p and q are continuous on an interval I : α < t < β containing
the point t = t0 , then there exists a unique function y that
satisfies the IVP
y 0 + p(t)y = g (t)
y (t0 ) = y0
for each t in I .
e.g. Find the largest interval in which the IVP
ty 0 + 2y = 4t 2 , y (1) = 2
has a unique solution.
Katie Walsh
Math 20D Final Exam Review
Theorem
Let f (t, y ) and ∂f /∂y be continuous for α < t < β, γ < y < δ
containing the point (t0 , y0 ). Then in some interval
t0 − h < t < t0 + h contained in α < t < β, there is a unique
solution to the IVP y 0 = f (t, y ), where y (t0 ) = y0
e.g. Does the theorem guarantee that y 0 = y 1/3 with the initial
condition y (0) = 0 has a unique solution?
No, since ∂f /∂y is undefined at y = 0. In fact, y = (2/3t)3/2 and
y = −(2/3t)3/2 are solutions to the IVP.
Katie Walsh
Math 20D Final Exam Review
Theorems for 2nd Order
Theorem
Consider the IVP
y 00 + p(t)y 0 + q(t)y = g (t), y (t0 ) = y0 , y 0 (t0 ) = y00
where p, q, g are continuous on an open interval I containing t0 .
Then there exists a unique solution y to this problem and the
solution exists throughout I .
Katie Walsh
Math 20D Final Exam Review
Theorems about the Wronskian
Theorem
Suppose that y1 and y2 are two solutions of the differential
equation
y 00 + p(t)y 0 + q(t)y = 0.
Then the family of solutions y = c1 y1 (t) + c2 y2 )t) with arbitrary
coefficients c1 and c2 includes every solutions to the above
differential equation if and only if there is a point t0 where the
Wronskian of y1 and y2 is not zero.
If y1 and y2 are both solutions to a homogeneous second order
differential equation and their Wronskian is nonzero for some point
t0 , we call y1 and y2 a fundamental solution set.
Katie Walsh
Math 20D Final Exam Review
Just because we can write every solutions as y = c1 y1 + c2 y2
doesn’t mean we can solve every initial value problem. (Why?
Because every solution can be expressed via infinitely many initial
conditions. Say the solution to the IVP is y (t) = 2t. This satisfies
the initial conditions y (0) = 0, y (1) = 2, y (7) = 14, . . . . Some of
these t0 may have W (y1 , y2 )(t0 ) = 0
Theorem
Let
y 00 + p(t)y 0 + q(t)y = 0,
have solutions y1 and y2 . The the IVP with initial conditions
y (t0 ) = y0 , y 0 (t0 ) = y00 has a solutions of the form y = c1 y1 + c2 y2
if and only if W (y1 , y2 )(t0 ) 6= 0
Katie Walsh
Math 20D Final Exam Review
Second Order, Linear, Homogeneous Differential Equations
with Constant Coefficients
ay 00 + by + c = 0
Characteristic Polynomial ar 2 + br + c = 0
Let r1 and r2 be the roots of the characteristic polynomial.
1
If r1 and r2 are real and distinct, then Let y1 = e r1 t and
y2 = e r2 t .
2
If r1 , r2 = λ + µi then y1 = e λt cosµt and y1 = e λt sinµt.
3
If r1 = r2 then y1 = e r1 t and y2 = te r1 t .
In each case, y1 and y2 form a fundamental solution set so the
general solution is y = c1 y1 + c2 y2 .
Katie Walsh
Math 20D Final Exam Review
Examples
1
y 00 − 2y 0 + y = 0
2
y 00 − y 0 − 6y = 0
3
y 00 − 10y 0 + 26y = 0
Katie Walsh
Math 20D Final Exam Review
Method of Undetermined Coefficients
Given the differential equation
y 00 + p(t)y 0 + q(t)y = g (t)
we can use the method of undetermined coefficients to find the
solutions. First we find the fundamental solution set to the
corresponding homogeneous equation. Then we guess the form of
the particular solutions.
1 If g (t) = A t n + A t n−1 + · · · + A
0
1
n−1 t + An , guess
B0 t n + B1 t n−1 + · · · + Bn−1 t + Bn
2 If g (t) = e αt , guess Ae αt
3 If g (t) = sin(αt) or g (t) = cos(αt), guess
Asin(αt) + Bcos(αt).
If we have a product of the above, guess the product (make sure to
have different constants or functions in front of the sin or cos
parts.) If we have a sum, break into different problems, finding a
g (t) for each term in the sum.
Katie Walsh
Math 20D Final Exam Review
Warning! If the guess is one of the homogeneous solutions,
multiply the guess by t. If it is still a solution to the homogeneous
solution (in the case of repeated roots), multiply by t again.
Katie Walsh
Math 20D Final Exam Review
What should we guess for the following:
1
2
3
−3te −t
(At + B)e −t
t 2 cos(2t)
(At 2 + Bt + C )cos(2t) + (Dt 2 + Et + F )sin(2t)
e t (sin(5t))
Ae t (cos(5t)) + Be t (sin(5t))
Once we have our guess, we plug it back in to the differential
equation. Then we can solve for the coefficients. Example:
y 00 − y = t 2 + 3e t
Katie Walsh
Math 20D Final Exam Review
Given
y 00 + p(t)y 0 + q(t)y = g (t)
we can use variation of parameters to solve. We first find y1 and
y2 the fundamental solution set. Then the particular is of the form
Z
Z
y2 (t)g (t)
y1 (t)g (t)
Y (t) = −y1
dt + y2 (t)
dt
W (y1 , y2 )
W (y1 , y2 )
As long as we remember our “+C ’s” this will in fact be the
general solution. Example: Consider
t 2 y 00 − 2y = 3t 2 − 1
Show that the corresponding homogeneous equation has
fundamental solution set y1 (t) = t 2 , y2 (t) = t −1 .
Use variation of parameters to find the general solution.
Katie Walsh
Math 20D Final Exam Review
New stuff since the first midterm
1
Systems of Linear Differential Equations
2
Series Solutions
3
Laplace Transforms
Katie Walsh
Math 20D Final Exam Review
Systems of Linear Differential Equations
1
2
Know how to find eigenvalues and eigenvectors of a 2x2
matrix.
2 −5
Example: Find eigenvalues and eigenvectors of A =
1 −2
Know how to find the general solution set to x0 = Ax in each
of the three cases below.
Case 1:
If Anxn has n-linearly independent
eigenvectors.
1
e.g. A has λ1 = −1 with v1 =
and λ2 = −4 with
2
1
v2 =
. What is the general solution?
−4
1
1
−t
x = c1
e + c2
e −4t .
2
−4
Katie Walsh
Math 20D Final Exam Review
Case 2
If A2x2 has complex conjugates as its eigenvalues.
1±i
Example: A has λ = 1 ± 2i with eigenvector v =
−4
1+i
Let x1 =
e 1+2i . Expand into real and imaginary parts.
−4
Then the general solution is
x = c1 (real part ofx1 ) + c2 imaginary part ofx1
Katie Walsh
Math 20D Final Exam Review
Case 3
If A2x2 has a repeated eigenvalue with only one linearly
independent set of eigenvectors, we need to find the generalized
eigenvector.
1 −1
Example:Let A =
. Then A has eigenvalue 2 with
1 3
1
as its only linearly independent eigenvector. Then we
−1
have to find the generalized eigenvector by solving (A − λI )w = v
i.e.
1
(A − 2I )w =
−1
The vector w will always have the form w = av + g where a is an
arbitrary constant. Then the general solution is
x = c1 ve λt + c2 (vte λt + ge λt )
Katie Walsh
Math 20D Final Exam Review
Theorems for Systems of First Order Linear Equations
Theorem
If x(1) , x(2) , · · · , x(n) are solutions to the system x0 = P(t)x on
α < t < β and W [x(1) , x(2) , · · · , x(n) ] 6= 0 for some point in that
interval, then x(1) , x(2) , · · · , x(n) form a fundamental set of
solutions for that interval.
Katie Walsh
Math 20D Final Exam Review
Series Solution
Given a second order linear homogeneous differential equation with
polynomial terms P(x)y 00 + Q(x)y 0 + R(x)y = 0 we can use series
solution.
Assume that the solution has the form
y=
∞
X
an (x − x0 )n
n=0
We need that x0 is an ordinary point of the differential equation,
i.e. P(x0 ) 6= 0. Then
y0 =
∞
X
nan x n−1
n=0
y 00 =
∞
X
n(n − 1)an x n−2
n=0
Katie Walsh
Math 20D Final Exam Review
Plug this in to the differential equation. Then we re-index as
necessary and add. We can then find a recurrence relation for the
coefficients. The first two coefficients are free, say a0 and a1 . If we
look at the terms with a0 we get one solution, the other solution
are the terms with a1 .
Example: (1 − x)y 00 + y = 0 with x0 = 0
Find the recurrence relation and the first four terms of each of the
two solutions.
Katie Walsh
Math 20D Final Exam Review
Theorem
If x0 is an ordinary point of P(x)y 00 + Q(x)y 0 + R(x)y = 0 and
Q(x)/P(x) and R(x)/P(x) have a Taylor Series at x0 then we get
a general solution
y=
∞
X
an (x − x0 )n = a0 y1 (x) + a1 y2 (x)
n=0
where a0 and a1 are arbitrary. The radius of convergence for y1
and y2 is at least as large as that of Q(x)/P(x) and R(x)/P(x).
Example: Find lower bound on radius of convergence of
(1 + x 3 )y 00 + 4xy 0 + y = 0 about x0 = 0 and x0 = 2.
Katie Walsh
Math 20D Final Exam Review
Euler Equations
Euler Equations have the form
ay 00 + by 0 + cy = 0
These have x0 = 0 as a singular point. We guess that the solution
y = x r . Plugging this in we get
x r (ar 2 + (b − a)r + c) = 0
We then find the roots of the quadratic equations.
Katie Walsh
Math 20D Final Exam Review
If the roots are real and distinct r = r1 , r2
y = c1 |x|r1 + c2 |x|r2
If the roots are repeated r = r1
y = c1 |x|r1 + c2 |x|r1 ln(x)
If the roots are complex r = λ ± µi
y = c1 |x|λ cos(µ ln |x|) + c2 |x|λ sin(µ ln |x|)
Katie Walsh
Math 20D Final Exam Review
Laplace Transform
Theorem
If f is a piecewise continuous on o ≤ t ≤ A for all A > 0 and
|f (t)| ≤ Ce at for t ≥ M then the Laplace transform of f is
Z ∞
L{f (t)} =
e −st f (t)dt
0
The Laplace transform is linear.
L{c1 f1 + c2 f2 } = c1 L{f1 } + c2 L{f2 }
Katie Walsh
Math 20D Final Exam Review
L{f 0 (t)} = sL{f (t)} − f (0)
L{f 00 (t)} = s 2 L{f (t)} − sf (0) − f 0 (0)
L{f (n) (t)} = s n L{f (t)} − s n−1 f (0) − · · · − sf (n−2) (0) − f (n−1) (0)
We can use these facts to solve differential equations. First take
Laplace transform of both sides using linearity. Then solve for
L{f (t)}. Then use table to find inverse Laplace Transform. You
may need to use partial fractions and complete the square.
Example: y 00 − 2y 0 + 2y = cos t with y (0) = 1 and y 0 (0) = 0
Katie Walsh
Math 20D Final Exam Review
The Unit Step Function (Heaviside Function)
For c > 0, we can define the unit stpe function at c.
0 :t<c
uc (t) =
1 :t≥c
Then
L{uc (t)} =
e −cs
s
and
L{uc (t)f (t − c)} = e −cs L{f }
We can use this to solve differential equations with discontinuos
forcing functions. E.g. y 00 + 3y 0 + 2y = u2 (t)
Katie Walsh
Math 20D Final Exam Review
The Dirac Delta Function
The dirac δ function is defined to satisfy the following properties:
δ(t − t0 ) = 0
Z ∞
t 6= t0
δ(t − t0 ) = 1
−∞
We can use Laplace Transforms to solve differential equations
involving δ(t − t0 )
L{δ(t − t0 )} = e −st0
Example: y 00 + 2y 0 + 2y = δ(t − π)
Katie Walsh
Math 20D Final Exam Review