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

1. Author: Daniel Albert May 25, 2018
www.dennisusa.com
1
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,
𝑑𝑣
𝑑𝑡
= 9.8 – 0.2t
• Order of differential equation: The highest derivative order
* Note: The highest derivative order ≠ the highest power. Y’’’ = Y(2)
≠ Y3
• Linear differential equation: the dependence (usually y) power is 1.
*Note: the power here ≠ derivative order!
For example: y” + 3y’ – 2t2
= 0 -> this is linear
y”’ – ty” + 1 = y2
-> This is NOT linear
(y’)y – x2
= 0 -> This is NOT linear

2. Author: Daniel Albert May 25, 2018
www.dennisusa.com
2
• Separable First Order Differential Equations
General form:
𝑑𝑦
𝑑𝑥
= 𝑓(𝑥, 𝑦)
Standard form:
𝑑𝑦
𝑑𝑥
= 𝑔( 𝑥)ℎ(𝑦)
• Reduction to separable form
𝑑𝑦
𝑑𝑥
= 𝑓 (
𝑦
𝑥
)
u =
𝑦
𝑥
=> 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

3. Author: Daniel Albert May 25, 2018
www.dennisusa.com
3
• Non-linear first order ordinary differential equation – Bernoulli equation
Standard form: y’ + P(x)y = Q(x)●yn
• Both sides divided by yn
=> Let Z = y1-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, x2
+ y2
= 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
www.dennisusa.com
4
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: ar2
+ br + c = 0
Let r1 and r2 be the roots of the characteristic polynomial.
Case 1: r1 and r2 are real distinct roots (means b2
– 4ac > 0), then y1 = er1x
, y2 = er2x
y = C1 er1x
+ C2 er2x
Case 2: r1 and r2 are repeated roots (means b2
– 4ac = 0), r1 = r2,

5. Author: Daniel Albert May 25, 2018
www.dennisusa.com
5
then y1 = er1x
, y2 = xer1x
y = C1 er1x
+ C2 xer1x
Case 3: r1 and r2 are complex roots (means b2
– 4ac < 0), r1 = α + βi,
r2 = α – βi, then y1 = eαx cos(βx) , y2 = y1 = eαx sin(βx)
y = C1 eαx cos(βx) + C2eαx sin(βx)
• Method of Undetermined Coefficients to solve 2nd
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
11x2
Ax2
+ Bx + C
82xn
Axn
+ Bxn-1
+ ……Z
-5e3x
Ae3x
sin(ax) or cos(ax) Asin(ax) + Bcos(ax)
−3te−t
(At+B)e−t
X2
cos(5x) (Ax2
+ Bx + C) cos(5x) + (Dx2
+ Ex +
F)sin(5x)
• Variation of parameter to solve 2nd
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

6. Author: Daniel Albert May 25, 2018
www.dennisusa.com
6
(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∫
𝑊1
𝑊
𝑅(𝑥)𝑑𝑥 + y2∫
𝑊2
𝑊
𝑅(𝑥)𝑑𝑥
• Euler-Cauchy Equation for homogeneous, Non-constant coefficients linear ODE
General form: ax2
y” + bxy’ =cy = 0
*Note: a, b, c are non-zero constant, x is not constant.
Use y = xm
• 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)

7. Author: Daniel Albert May 25, 2018
www.dennisusa.com
7
Solutions y1 = xm1, y2 = xm2.
y = C1 xm1 + C2xm2
• Case 2: m1 and m2 are real, repeat roots (m1 = m2)
y1 = xm1, y2 = xm1ln(x)
y = C1 xm1 + C2xm1ln(x)
• Case 3: m1 and m2 are distinct, complex roots.
m = α  βi
y1 = x α + βi, y2 = xα – βi
Y = C1xαcos(βlnx) + C2xαsin(βlnx)
= xα[C1cos(βlnx) + C2sin(βlnx)]
Chapter 3: Higher order linear ODE
Similar to 2nd oreder linear ODE to find solutions.
General form: axn
y(n)
+ bxn-1
y(n-1)
+ cxn-2
y(n-2)
+ …… + zy = 0 (Homogeneous)
axn
y(n)
+ bxn-1
y(n-1)
+ cxn-2
y(n-2)
+ …… + zy = R(x) (non-homogeneous)

8. Author: Daniel Albert May 25, 2018
www.dennisusa.com
8
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) -> 1st order ODE
x2’ = f2(x1, x2 ,,,,xn) -> 2nd order ODE
x3’ = f3(x1, x2 ,,,,xn) -> 3rd order ODE
xn’ = fn(x1, x2 ,,,,xn) -> nth 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)

9. Author: Daniel Albert May 25, 2018
www.dennisusa.com
9
(1) * -(D-1) + (2) => -(D-1) (D-1) x1 – x1 = 0 (Note: eliminate x2)
=> (D2 – 2D + 2) x1 = 0, x1 = y
=> (D2 – 2D + 2) y = 0
=> y” – 2y’ + 2y = 0.
• Linear system of 1st 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

10. Author: Daniel Albert May 25, 2018
www.dennisusa.com
10
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 = C1eαt(acos(βt) - bsin(βt)) + C2eαt(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