Applied Mathematics II Revision Note
John NG
Ordinary Differential Equation (Part one)

”± Order and Degree

Order”GHighest derivative
Degree”GHighest power of the highest derivative


Order = 4
Degree = 2

”± First Order D.E.

Variable Separable
Linear First Order
Solving Technique

Separate x and y
from both sides completely.

Let , then


By integrating factor

and the solution is

Further technique”GInspection (e.g. you should know xdy/dx + y is d(xy)/dx immediately) and Substitution

”± Typical Examples of First Order D.E.

Variable Separable”Ge.g. Rate of change of y is directly proportional to y.

=> =>
If initial condition t = 0, y = C0 is given, we have

Homogeneous”Ge.g. Fraction in Linear Polynomials (Can be converted into homogeneous type.)

Type A”@A:B = D:E Coeffs. in proportion Type B”@A:B ”Ś D:E Coeffs. not in proportion
Technique”GLet u = Ax + By”Aconvert the DE involves u and x only. Technique”GLet x = u + h”Ay = v + k”Afind h, k to convert the DE into DE involves u and v such that no constant terms occur in the fraction.
Example Example
Solution Solution
Let , then
and the DE will be


Let , then ,and the DE will be
, to make homo., set
, having h = 2”Ak = ”Š1, hence
and the DE will be homogeneous,

Let , then , and DE will be

”@(Partial fraction)

Linear First Order”Ge.g. Bernoull's Equation after suitable transformation.

Solve by the transformation of .

If , then , the DE will be

=> (Linear First Order)
Integrating factor:
The solution is


”± Linear Second Order D.E.

Homogeneous”G”@”@ (a, b, c are constants with a is non-zero)

Non-Homogeneous”G (f(x) is non-zero function)

”± Solving Homogeneous Linear Second Order D.E.

Solve .

Step 1”GSolve the auxiliary equation

Hence = ,
Step 2”GWrite down the solution according to the following three cases.
Distinct real roots ,
Equal roots = =
Complex roots , =

”± Solving Non-Homogeneous Linear Second Order D.E.

Solve .

Step 1”GSolve the homogeneous DE. Obtain the solution yc.
Solve by the method mentioned above and obtain the solution yc.
Step 2”GGuess a particular solution yp by the method of undetermined coefficients.
By observing f(x), we can guess out yp easily in the following ways:
Sine and Cosine
Step 3”GModification for yp ?
If there is an expression in the guessing yp occurs in yc, just multiply the guessing yp by x. Continue this "multiplication process" until no term in yp occurs in yc.
Step 4”GSubstitute yp into the original DE to determine the coefficients in the yp.
Step 5”GThe general solution to the DE is y = yc + yp.

”± Typical Examples of Linear Second Order D.E.

Example 1”GSolve .

Step 1”GSolve
=> = 1, 2 => yc =
Step 2”GGuess yp.
Since R.H.S. = ex, we guess yp = Cex.
Step 3”GModify yp.
Since ex occurs in yc, we need to modify yp as yp = Cxex.
Step 4”GDetermine coefficients in yp.
Let yp = Cxex, then and ,
substitute them back to the original DE,
, hence C = 1, yield yp = xex.
Step 5”GWrite down the general solution y = yc + yp.

Example 2”GSolve .

Step 1”GSolve
Step 2”GGuess yp.
Beware! R.H.S. involves cos2x NOT cosx. We need to use . Hence the DE will be , thus .
Step 3”GModify yp.
Not necessary.
Step 4”GDetermine coefficients in yp.
, , sub. back to the original DE, we have

=>, and =>
Step 5”GWrite down the general solution y = yc + yp.

Example 3”GSolve . (Hint: let .)

Try to convert the independent variable t into u through the transformation .
Aim at writing , into something involves , . First of all, , hence . Now,


Substitute back to the original equation, we have

--- (*)

It's very easy to obtain . And it's trivial to guess , consequentially,
and . Substitute into (*), , yield , thus
; therefore the general solution to the original DE is , i.e.

Applied Mathematics II Revision Note