”± Order and Degree
Order”GHighest derivative
Degree”GHighest power of the highest derivative
Example
”± First Order D.E.
Variable
Separable

Homogeneous

Linear
First Order


Pattern  
Solving Technique 
Separate x and y
from both sides completely. 
Let , then
Yield 
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.
Solution 
=> =>
If initial condition t = 0, y = C_{0} 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 hence
i.e. 
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.
Solution 
If
, then
, the DE will be
=> (Linear First Order) Integrating factor: The solution is Hence 
”± Linear Second Order D.E.
Homogeneous”G”@”@
(a, b, c are constants with a is nonzero)
NonHomogeneous”G
(f(x) is nonzero function)
”± Solving Homogeneous Linear Second Order D.E.
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 ,
=

Solution”G 
Solution”G or 
Solution”G or 
”± Solving NonHomogeneous Linear Second Order D.E.
Step 1”GSolve the homogeneous DE. Obtain the solution y_{c}. 
Step 2”GGuess a particular solution y_{p} by the method of undetermined coefficients. 
By observing
f(x), we can guess out y_{p} easily in the following
ways:

f(x)

y_{p}


Exponential


Polynomials


Sine and Cosine

or

Step 3”GModification for y_{p} ? 
If there is an expression in the guessing y_{p} occurs in y_{c}, just multiply the guessing y_{p} by x. Continue this "multiplication process" until no term in y_{p} occurs in y_{c}. 
Step 4”GSubstitute y_{p} into the original DE to determine the coefficients in the y_{p}. 
Step 5”GThe general solution to the DE is y = y_{c} + y_{p}. 
”± Typical Examples of Linear Second Order D.E.
Example 1”GSolve .
Step 1”GSolve 
=> =
1, 2 => y_{c} =

Step 2”GGuess y_{p}. 
Since R.H.S.
= e^{x}, we guess y_{p} = Ce^{x}.

Step 3”GModify y_{p}. 
Since e^{x}
occurs in y_{c}, we need to modify y_{p}
as y_{p} = Cxe^{x.}

Step 4”GDetermine coefficients in y_{p}. 
Let y_{p}
= Cxe^{x}, then
and ,
substitute them back to the original DE, , hence C = 1, yield y_{p} = xe^{x.} 
Step 5”GWrite down the general solution y = y_{c} + y_{p}. 
i.e.

Example 2”GSolve .
Step 1”GSolve 
Step 2”GGuess y_{p}. 
Beware! R.H.S.
involves cos^{2}x NOT cosx. We need to use
. Hence the DE will be ,
thus .

Step 3”GModify y_{p}. 
Not necessary^{.}

Step 4”GDetermine coefficients in y_{p}. 
, ,
sub. back to the original DE, we have =>, and => 
Step 5”GWrite down the general solution y = y_{c} + y_{p}. 
i.e.

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, and  (*) 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. 