### Applied Differential Equations

```Differential Equations Study Guide1
First Order Equations
dy
= f (x, y)
dx
Initial Value Problem: y 0 = f (x, y), y(x0 ) = y0
General Form of ODE:
(1)
(2)
Linear Equations
General Form: y 0 + p(x)y = f (x)
(3)
(4)
(5)
(6)
Method for Solving Exact Equations:
R
1. Let φ = M (x, y)dx + h(y)
R
p(x)dx
Integrating Factor: µ(x) = e
d
(µ(x)y) = µ(x)f (x)
=⇒
dx
Z
1
General Solution: y =
µ(x)f (x)dx + C
µ(x)
Homogeneous Equations
∂φ
= N (x, y)
∂y
3. Simplify and solve for h(y).
2. Set
4. Substitute the result for h(y) in the expression for φ from step
1 and then set φ = 0. This is the solution.
Alternatively:
R
1. Let φ = N (x, y)dy + g(x)
∂φ
= M (x, y)
∂x
3. Simplify and solve for g(x).
2. Set
(7)
(8)
General Form: y 0 = f (y/x)
Substitution: y = zx
4. Substitute the result for g(x) in the expression for φ from step
1 and then set φ = 0. This is the solution.
=⇒ y 0 = z + xz 0
(9)
The result is always separable in z:
dx
dz
=
f (z) − z
x
(10)
Integrating Factors
Case 1: If P (x, y) depends only on x, where
Bernoulli Equations
(11)
(12)
0
General Form: y + p(x)y = q(x)y
n
Substitution: z = y 1−n
(18)
P (x, y) =
R
M y − Nx
=⇒ µ(y) = e P (x)dx
N
then
The result is always linear in z:
(13)
z 0 + (1 − n)p(x)z = (1 − n)q(x)
(19)
µ(x)M (x, y)dx + µ(x)N (x, y)dy = 0
is exact.
Exact Equations
Case 2: If Q(x, y) depends only on y, where
(14)
(20)
(15)
(16)
(17)
General Form: M (x, y)dx + N (x, y)dy = 0
∂M
∂N
Text for Exactness:
=
∂y
∂x
Solution: φ = C where
∂φ
∂φ
M=
and N =
∂x
∂y
Q(x, y) =
R
Nx − M y
=⇒ µ(y) = e Q(y)dy
M
Then
(21)
µ(y)M (x, y)dx + µ(y)N (x, y)dy = 0
is exact.
1 « 2014
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Second Order Linear Equations
General Form of the Equation
Heuristics for Undetermined Coefficients
(Trial and Error)
(22)
General Form: a(t)y 00 + b(t)y 0 + c(t)y = g(t)
(23)
Homogeneous: a(t)y 00 + b(t)y 0 + c(t)y = 0
(24)
Standard Form: y 00 + p(t)y 0 + q(t)y = f (t)
If f (t) =
Pn (t)
Pn (t)eat
Pn (t)eat sin bt
or Pn (t)eat cos bt
then guess that a particular solution yp =
ts (A0 + A1 t + · · · + An tn )
ts (A0 + A1 t + · · · + An tn )eat
ts eat [(A0 + A1 t + · · · + An tn ) cos bt
+(A0 + A1 t + · · · + An tn ) sin bt]
The general solution of (22) or (24) is
(25)
y = C1 y1 (t) + C2 y2 (t) + yp (t)
Method of Reduction of Order
where y1 (t) and y2 (t) are linearly independent solutions of (23).
Linear Independence and The Wronskian
When solving (23), given y1 , then y2 can be found by solving
(36)
y1 y20 − y10 y2 = Ce−
R
p(t)dt
The solution is given by
Two functions f (x) and g(x) are linearly dependent if there
Z − R p(x)dx
e
dx
exist numbers a and b, not both zero, such that af (x) + bg(x) = 0
(37)
y2 = y1
2
for all x. If no such numbers exist then they are linearly indey1 (x)
pendent.
If y1 and y2 are two solutions of (23) then
(26)
(27)
Wronskian: W (t) =
Method of Variation of Parameters
y1 (t)y20 (t) −
R
− p(t)dt
y10 (t)y2 (t)
Abel’s Formula: W (t) = Ce
and the following are all equivalent:
1. {y1 , y2 } are linearly independent.
2. {y1 , y2 } are a fundamental set of solutions.
If y1 (t) and y2 (t) are a fundamental set of solutions to (23) then
a particular solution to (24) is
Z
Z
y2 (t)f (t)
y1 (t)f (t)
(38)
yP (t) = −y1 (t)
dt + y2 (t)
dt
W (t)
W (t)
Cauchy-Euler Equation
3. W (y1 , y2 )(t0 ) 6= 0 at some point t0 .
4. W (y1 , y2 )(t) 6= 0 for all t.
ODE: ax2 y 00 + bxy 0 + cy = 0
(40)
Auxilliary Equation: ar(r − 1) + br + c = 0
The solutions of (39) depend on the roots r1,2 of (40):
Initial Value Problem
(28)
(39)
(41)
 00
 y + p(t)y 0 + q(t)y = 0
y(t0 ) = y0
 0
y (t0 ) = y1
Real Roots: y = C1 xr1 + C2 xr2
(42) Repeated Root: y = C1 xr + C2 xr ln x
(43)
Complex: y = xα [C1 cos(β ln x) + C2 sin(β ln x)]
In (43) r1,2 = α ± iβ, where α,β ∈ R
Linear Equation: Constant Coefficients
Series Solutions
(29)
(30)
(31)
(32)
00
0
Homogeneous: ay + by + cy = 0
Non-homogeneous: ay 00 + by 0 + cy = g(t)
Characteristic Equation: ar2 + br + c = 0
√
−b ± b2 − 4ac
2a
(44)
If x0 is a regular point of (44) then
(45)
Real Roots(r1 6= r2 ) : yh = C1 er1 t + C2 er2 t
(34)
Repeated(r1 = r2 ) : yh = (C1 + C2 t)er1 t
(35)
Complex(r = α ± iβ) : yH = eαt (C1 cos βt + C2 sin βt)
y1 (t) = (x − x0 )n
∞
X
ak (x − xk )k
k=0
The solution of (29) is given by:
(33)
(x − x0 )2 y 00 + (x − x0 )p(x)y 0 + q(x)y = 0
At a Regular Singular Point x0 :
(46)
(47)
Indicial Equation: r2 + (p(0) − 1)r + q(0) = 0
∞
X
First Solution: y1 = (x − x0 )r1
ak (x − xk )k
k=0
The solution of (30) is y = yp + yh where yh is given by (33)
through (35) and yp is found by undetermined coefficients or Where r1 is the larger real root if both roots of (46) are real or
reduction of order.
either root if the solutions are complex.
```