Differential Equations Basic
Differential Equations Basic
Differential Equations Basic
• Differential Equations
– References
– General Concepts and Definitions
– Method of Separation of Variables
– Method of Transformation of Variables
∗ Homogeneous Equations
– Exact DE and Integrating Factors
References
GLA Notes
dn y dn−1 y
f (x) = an (x) + an−1 (x) + . . . a0 (x)y (Linear ODE)
dxn dxn−1
Homogeneous: has zero as a solution. Otherwise, nonhomogeneous
Partial Differential Equation (PDE): equation with more than one inde-
pendent variable involving these as well as various partial derivatives w/ respect
to the variables
When solving ODEs such that the derivative is brought to y(x), or simply y,
there will be n conditions required to find the n constants of an nth-order ODE.
Initial Value Problem: DE is required to satisfy conditions on the dependent
variable and its derivatives specified at one value of the independent variable
• Eg. Equation of motion can be solved at time zero
Boundary Value Problem: DE is required to satisfy conditions on the de-
pendent variable and its derivatives at 2+ values of the independent variable
Existence and Uniqueness Theorum: suggest that existence and uniqueness
of a solution is guaranteed when the specified conditions hold. Even specified
conditions are not all satisfied, there may still exist a unique solution
1
Method of Separation of Variables
dy
Variable separable: a first order ODE dx = F (x, y) such that F(x,y) can be
separated into f (x) · φ(y)
Procedure:
1. Case 1: If φ(y) =6 0, separate terms to make it integrable. Integrate for
GS.
2. Case 2: If φ(y) = 0, solve for the roots of this equation. These will be
solutions to the differential equation
When it is hard to completely find general solution, it is okay to write a singular
solution
y = xv
dy dv
=v+x
dx dx
dv
v+x = f (v)
dx
dv
x = f (v) − v
dx
Procedure for analysis:
1. Case 1: If f (v) − v = 0, it is known that y = v0 x. Thus, v0 is the solution
of f (v0 ) − v0 = 0
2. Case 2: If f (v) − v 6= 0, the below general solution is used
Z Z
dv dx
= +C
f (v) − v x
2
Particular Solution: solution found from general solution when plugging in
givens from the question
Singular Solution: solution found from testing case one when it equals zero.
It can be expressed aside from the general solution
If the question consists of two linear functions that are non-homogeneous, the
axes need to be changed. This can be done according as following
1. Using Gaussian Elimination, find POI (h,k)
2. Switch axes by letting x = X + h and y = Y + k
3. At the end of the question, revert to the original coordinate system
3
∂M
∂y − ∂N
∂x
Case 1: if N contains x only, the below is true
∂M ∂N
Z
∂y − ∂x
µ(x) = exp ( dx)
N
∂M
∂y − ∂N
∂x
Case 2: if M contains y only, the below is true
∂N ∂M
Z
∂x − ∂y
µ(y) = exp ( dy)
M