Differential equation study guide for exam (formula sheet)

This is the study guide for differential equation for engineering math analysis.

1. Author: Daniel Albert May 25, 2018
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Differential Equation and Linear Algebra Exam Review (The Cheat Sheet)
Chapter 1: First order linear ODE
• Differential equation: A function relationship between that function and its
derivative.
F (y, y”, y”, y’’’, y’’’’,,,,y
(n)
) = 0
• Ordinary differential equation (ODE): differential equation that depends on a
single variable.
For example,
????????????
????????????
= 9.8 – 0.2t
• Order of differential equation: The highest derivative order
* Note: The highest derivative order ≠ the highest power. Y’’’ = Y
(2)
≠ Y
3

• Linear differential equation: the dependence (usually y) power is 1.
*Note: the power here ≠ derivative order!
For example: y” + 3y’ – 2t
2
= 0 -> this is linear
y”’ – ty” + 1 = y
2
-> This is NOT linear
(y’)y – x
2
= 0 -> This is NOT linear

2. Author: Daniel Albert May 25, 2018
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• Separable First Order Differential Equations
General form:
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??????�
=�(�,�)
Standard form:
??????�
??????�
=�(�)ℎ(�)
• Reduction to separable form
??????�
??????�
=�(
�
�
)
u =
�
�
=> y = ux => y’ = xu’ + u
• Exact equation
Standard form: M(x,y)dx + N(x,y)dy = 0
Non-exact equation (use integrating factor)
μ(x) = e
∫ (dM/dy – dN/dx) / N dx
μ(y) = e
∫ (dN/dx - dM/dy) / M dy

• Linear first rrder ordinary differential equations
Standard form: y’ + P(x)y = Q(x)
μ(x) = e
∫ p(x) dx

y = [1/ μ(x)] * ∫ μ(x)●Q(x) dx

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• Non-linear first order ordinary differential equation – Bernoulli equation
Standard form: y’ + P(x)y = Q(x)●y
n

• Both sides divided by y
n

=> Let Z = y
1-n
, -> Z’ = (1-n) y
-n
●y’
=>New linear equation in Z: Z’ + (1-n)P(x)Z = (1-n)Q(x)
• Find orthogonal trajectories to the family of curves
For example, x
2
+ y
2
= C (a)
Step 1: Find slope m1 = y’
Take derivative of both side of (1): 2x + 2yy’ = 0
=> m1 = y’, y’ = -x/y
Step 2: m1 ● m2 = -1 (dot product of both slopes is -1 for orthogonal trajectories)
Step 3: m2 = -1 / m1
=> m2 = y/x => dy/dx = y/x (separable equation)
Solve for family curve get: ln|y| – ln|x| = k

4. Author: Daniel Albert May 25, 2018
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Chapter 2: Second order linear ODE
• Standard form: y” + P(x)y’ + Q(x)y = 0 (homogeneous)
o Principle of superposition solution: Y = C1y1(x) + C2y2(x)
*Note: y1 and y2 are linearly independent solutions.
• To find y1 and y2:
Method 1: Given y1, use Lagrange reduction order to find y2.
y2(x) = u(x)●y1(x)
Method 2: Use characteristic equation as introduced below.
Method 3: Undetermined coefficient.
Method 4: Euler-cauchy formula.
Method 5: Variation of parameter.
• Second order linear, homogeneous ODE with constant coefficients.
General form: ay” + by’ + cy = 0.
Characteristic (AKA “Auxiliary”) equation: ar
2
+ br + c = 0
Let r1 and r2 be the roots of the characteristic polynomial.
Case 1: r1 and r2 are real distinct roots (means b
2
– 4ac > 0), then y1 = e
r1x
, y2 = e
r2x

y = C1 e
r1x
+ C2 e
r2x
Case 2: r1 and r2 are repeated roots (means b
2
– 4ac = 0), r1 = r2,

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then y1 = e
r1x
, y2 = xe
r1x

y = C1 e
r1x
+ C2 xe
r1x

Case 3: r1 and r2 are complex roots (means b
2
– 4ac < 0), r1 = α + βi,
r2 = α – βi, then y1 = e
αx
cos(βx) , y2 = y1 = e
αx
sin(βx)
y = C1 e
αx
cos(βx) + C2e
αx
sin(βx)
• Method of Undetermined Coefficients to solve 2
nd
order Non-homogeneous linear
ODE
Standard form: y” + P(x)y’ + Q(x)y = R(x) (Non-homogeneous)
First we find the fundamental solution set to the corresponding homogeneous
equation. Then we guess the form of the particular solutions based on the pattern of
R(x):
R(x) Suggested yp
3x Ax + B
11x
2
Ax
2
+ Bx + C
82x
n
Ax
n
+ Bx
n-1
+ ……Z
-5e
3x
Ae
3x
sin(ax) or cos(ax) Asin(ax) + Bcos(ax)
−3te
−t
(At+B)e
−t

X
2
cos(5x) (Ax
2
+ Bx + C) cos(5x) + (Dx
2
+ Ex +
F)sin(5x)

• Variation of parameter to solve 2
nd
order non-homogeneous linear ODE
Suppose y1(x) and y2(x) are independent solution of the homogeneous equation. Let
W(y1, y2) be the Wrongskian of y1, y2. Note that y1 and y2 are linearly independent

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(means that W(y1, y2) ≠0). Different books define W1 and W2 differently, but the
result for general solution y should be the same.
W(y1, y2) = y1 y2 = y1y2’ – y2y1’, W1 = 0 y2 = - y2
y1’ y2’ 1 y2’,
W2 = y1 0
y1’ 1
Note: When using variation of parameter to solve ODE, the equation should change to
standard form: y” + P(x)y’ + Q(x)y = R(x). The coefficient for y” should be 1 in order to use
the formula.
Particular solution: Yp = y1∫
??????1
??????
??????(&#3627408485;)??????&#3627408485; + y2∫
??????2
??????
??????(&#3627408485;)??????&#3627408485;

• Euler-Cauchy Equation for homogeneous, Non-constant coefficients linear ODE
General form: ax
2
y” + bxy’ =cy = 0
*Note: a, b, c are non-zero constant, x is not constant.
Use y = x
m

• Characteristic equation for second order non-constant coefficients using Euler-Cauchy
method: am(m-1) + bm + c = 0
Find m1, m2.
• Case 1: m1 and m2 are real distinct roots (m1≠m2)

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Solutions y1 = x
m1
, y2 = x
m2
.
y = C1 x
m1
+ C2x
m2
•
Case 2: m1 and m2 are real, repeat roots (m1 = m2)
y1 = x
m1
, y2 = x
m1
ln(x)
y = C1 x
m1
+ C2x
m1
ln(x)

•
Case 3: m1 and m2 are distinct, complex roots.
m = α  βi
y1 = x
α + βi
, y2 = x
α – βi
Y = C1x
α
cos(βlnx) + C2x
α
sin(βlnx)
= x
α
[C1cos(βlnx) + C2sin(βlnx)]

Chapter 3: Higher order linear ODE
Similar to 2
nd
oreder linear ODE to find solutions.
General form: ax
n
y
(n)
+ bx
n-1
y
(n-1)
+ cx
n-2
y
(n-2)
+ …… + zy = 0 (Homogeneous)
ax
n
y
(n)
+ bx
n-1
y
(n-1)
+ cx
n-2
y
(n-2)
+ …… + zy = R(x) (non-homogeneous)

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Note: the number of solution is the same number as the highest order of
derivative. For example, y
(5)
has the highest order derivative as 5, therefore
there should be 5 solutions – y1, y2, y3, y4, y5.
Chapter 4: System of ODE
General form: F (x, y, y’, y”, y”’,,,,, y
(n)
) = 0
x1’ = f1(x1, x2 ,,,,xn) -> 1
st
order ODE
x2’ = f2(x1, x2 ,,,,xn) -> 2
nd
order ODE
x3’ = f3(x1, x2 ,,,,xn) -> 3
rd
order ODE
xn’ = fn(x1, x2 ,,,,xn) -> n
th
order ODE
f(x) = a x1+ bx2 +,,,cxn
Note: x1, x2 ,,,,xn are variables.
• Transform system to linear form (elimination method)
Let x1 = y, x2= y’, x1’ = y’, x2’ = y”
Example:

=> =>
X1’ = x1 – x2
X2’ = x1 + x2
Dx1 = x1 – x2
Dx2 = x1 + x2
(D-1) x1 + x2 = 0 (1)
- x1 + (D-1) x2 = 0 (2)

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(1) * -(D-1) + (2) => -(D-1) (D-1) x1 – x1 = 0 (Note: eliminate x2)
=> (D
2
– 2D + 2) x1 = 0, x1 = y
=> (D
2
– 2D + 2) y = 0
=> y” – 2y’ + 2y = 0.
• Linear system of 1
st
order ODE with constant coefficient (Homogeneous)
x1’ = a11 x1 + a12 x2 + ,,,,,, + a1n xn
x2’ = a21 x1 + a22 x2 + ,,,,,, + a2n xn
x2’ = an1 x1 + an x2 + ,,,,,, + ann xn


A =

Note: x1, x2 ,,,,xn are variables, a11, a12, ,,,, ann are non-zero constant.
Standard form: X’ = AX (homogeneous)
General solution: X = C1X1 + C2X2 + ,,,,,, + CnXn
• Case 1: A has real distinct eigenvalues.
X’ = AX, then find eigenvalue |A - I|=0, v is eigenvector.
a1 a2 a3
b1 b2 b3
c1 c2 c3
dn1 dn2 dn3

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x1 = v1e
1t

x2 = v1e
2t
xn = vne
nt
General solution X = C1 v1e
1t
+ C2v1e
2t
• Case 2: A has complex eigenvalues.
X’ = AX
= α  βi
Use = α + βi to find eigenvector v = a + bi, (Note: a, b here are vectors)
General solution X = C1e
αt
(acos(βt) - bsin(βt)) + C2e
αt
(asin(βt) - bcos(βt))
• Case 3: Repeated roots
Choose a vector A●a ≠ ●a, (vector a ≠ v)
b = A●a - ●a
General solution X = C1 be
t
+ C2(bt + a)e
t