2nd Order Differential Eqns

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2nd Order Differential Equations
Type 1 click for link
Type 2 click for link
Exceptions to the Particular Integral Rule
1st order equations which can be solved using the 2nd order method
Equations which contain a
equations.
d2 y
dx 2
term are called 2nd order differential
Type 1
d2 y
dy
a 2 b
 cy  0
dx
dx
Step1 Define and solve the auxiliary equation
al2 + bl + c = 0
Step 2 Write down the solution using the table below
Roots of auxiliary equation
Solution
Real and different
 and 
y = Aex + Bex
Real and equal (Repeated)

y = (Ax + B)ex
Complex roots   i 
y = ex(Asinx + Bcosx)
These solutions are called the complementary function
The complementary function is the solution to the equation
d2 y
dy
a 2 b
 cy  0
dx
dx
Ex1
Solve
d2 y
dy

4
 3y  0
dx 2
dx
Auxiliary equation
1l2 – 4l + 3 = 0
Solving
(l  3)(l  1)  0
Ex2
Solve
i.e
 and 
y = Ae3x + Be1x
Complementary function solution
Roots of auxiliary equation
Real and different
Real and equal (Repeated)
l = 3 or 1
 and 

Solution
y = Aex + Bex
y = (Ax + B)ex
d2 y
dy

4
 4y  0
dx 2
dx
Auxiliary equation
1l2 – 4l + 4 = 0
Solving
(l  2)(l  2) = 0
Complementary function solution
l=2
y = (Ax + B)e2x
i.e  repeated root
Finding the two constants
A 1st order differential equation contains one constant and a 2nd order contains
two.
Hence we need two boundary conditions.
Ex1
d2 y
dy

4
 4y  0
2
dx
dx
Given that: when x = 0
y=8
Auxiliary equation
l2 – 4l + 4 = 0
Solving
(l  2)(l  2) = 0
Complementary function solution
Now apply the boundary conditions
and when
dy
1
dx
l=2
x=0
i.e  repeated root
y = (Ax + B)e2x
y = (Ax + B)e2x
a) Substitute x = 0 and y = 8
8 = Be0
B=8
b) Substitute
dy
= 1 when x = 0
dx
dy
= (Ax + B)2e2x + e2x(A)
dx
Using the product rule: 1stgrad 2nd + 2ndgrad 1st
1 = B(2e0) + Ae0
Substituting x = 0 and = 1
1 = 2B + A
Substitute B = 8 so A = –15
The complete solution is therefore y = (Ax + B)e2x with A = –15 and B = 8
y = (–15x + 8)e2x
Type 2
d2 y
dy
If the equation is of the form a 2  b
 cy  f(x)
dx
dx
then the solution is made up of the sum of the complementary function and a
particular integral.
The particular integral is determined using the table below.
The particular integral is an expression which satisfies the differential equation
when it is substituted in.
f(x)
linear polynomial
ax + b
exponential function
aekx
trig fn acospx or bsinpx or both
Particular integral
Cx + D
Cekx
Ccospx + Dsinpx
F(x)
Ex1
d2 y
dy
4 2 4
 y  3x  4
dx
dx
Solve
4l2 + 4l + 1 = 0
Auxiliary equation
Solving
(2l  1)(2l  1) = 0
l = – i.e  repeated root
Complementary function solution
f(x)
linear polynomial
ax + b
exponential function
aekx
trig fn acospx or bsinpx or both
y = (Ax + B)e–x
Particular integral
Cx + D
Cekx
Ccospx + Dsinpx
F(x)
Particular integral is of the form y = Cx + D.
This must satisfy the differential equation
y = Cx + D
dy
C
dx
d2 y
0
2
dx
d2 y
dy
4 2 4
 y  3x  4
dx
dx
d2 y
dy
4 2 4
 y  3x  4
dx
dx
y = Cx + D
Substituting into the differential equation gives:
4(0) + 4C + Cx + D = 3x + 4
Comparing the coefficients
x`s
no.s
C=3
4C + D = 4
D = –8
Thus the particular solution is y = 3x – 8
The complete general solution is given by:
y = Complementary function + Particular integral
y = (Ax + B)e–x + 3x – 8
dy
C
dx
d2 y
0
2
dx
Ex2
Solve
d2 y dy
2 2
 3y  e2x
dx
dx
Auxiliary equation
Solving
2l2 – 1l – 3 = 0
(2l  3)(l  1)  0
l=
Complementary function solution
f(x)
linear polynomial
exponential function
trig fn
ax + b
aekx
acospx or bsinpx or both
y = Ae
or –1
x
+ Be–1x
Particular integral F(x)
Cx + D
Cekx where k has the same value as
the f(x) term
Ccospx + Dsinpx
Particular integral is of the form y = Ce2x.
This must satisfy the differential equation
d2 y dy
2 2
 3y  e2x
dx
dx
d2 y dy
2 2
 3y  e2x
dx
dx
y=
Ce2x
dy
 2Ce2x
dx
d2 y
 4Ce2x
2
dx
Substituting into the differential equation gives:
24Ce2x – 2Ce2x – 3Ce2x = e2x
Comparing the coefficients
8C – 2C – 3C = 1
e2x`s
C = 1/3
Thus the particular solution is y = 1/3e2x
The complete general solution is given by:
y = Complementary function + Particular integral
y = Ae
x
+ Be–1x + 1/3e2x
Solve
d2 y
1 2  4y  3 sin x  4 cos x
dx
dy
Here the
term is missing so the auxiliary equation becomes
dx
1l2 + 4 = 0
l = (0 + 2i)
as –4 = 2i
Complex roots
Complementary function solution
y = e0x(Asin2x + Bcos2x) = Asin2x + Bcos2x
f(x)
linear polynomial
ax + b
exponential function
aekx
trig fn acospx or bsinpx or both
Particular integral
Cx + D
Cekx
Ccospx + Dsinpx
Particular integral is of the form y = Ccosx + Dsinx.
y = Ccosx + Dsinx
dy
 C sin x  Dcos x
dx
d2 y
 Ccos x  Dsin x
2
dx
F(x)
d2 y
1 2  4y  3 sin x  4 cos x
dx
dy
 C sin x  Dcos x
dx
y = Ccosx + Dsinx
d2 y
 Ccos x  Dsin x
2
dx
Substituting into the differential equation gives:
–Ccosx – Dsinx + 4(Ccosx + Dsinx) = 3sinx + 4cosx
3Ccosx + 3Dsinx = 3sinx + 4cosx
Equating coefficients of sinx and cosx
3D = 3
D=1
3C = 4
C=
The complete general solution is given by:
y = Complementary function + Particular integral
y = Asin2x + Bcos2x + 1sinx +
cosx
Exceptions to the Particular Integral Rule
If the particular integral is the same as one of the terms in the complementary
function then use
y = xf(x) as the particular integral.
d2 y dy
2 2
 3y  e x
dx
dx
2l2 – 1l – 3 = 0
(2l - 3)(l + 1) = 0
l=
or
–1
Complementary function solution
Auxiliary equation
Solving
y = Ae
x
+ Be–1x
As f(x) = e–x is contained in the complementary function solution use
y = xCe–x
as the P.I.
d2 y dy
2 2
 3y  e x
dx
dx
Particular integral is
y = Cxe–x
C.F.
y = xCe–x
y = Ae
x
+ Be–1x
P.I
y = xCe–x
as the f(x) term contains e–x as does the C.F
dy
Use the product rule to find
dx
and
d2 y
dx 2
dy
 Cx  e  x  e  x  C = Ce−x (−x + 1)
dx
d2 y
 Ce x ( 1)  (  x  1)( Ce x )
2
dx
Substituting into the differential equation gives:
2 Ce x ( 1)  (  x  1)( Ce  x )  Ce-x (-x +1)  3Cxe  x  e  x
2 Ce x ( 1)  (  x  1)( Ce  x )  Ce-x (-x +1)  3Cxe  x  e  x
Comparing the coefficients
–2C – 2C –1C
e–x`s
Thus the particular solution is
=1
C=–
y = – xe–x
The complete general solution is given by:
y = Complementary function + Particular integral
y = Ae
x
+ Be–x –
xe–x
Summary
Complementary
function contains
Aex
f(x) ie RHS of =
Cex
Particular Integral
Dxex
Acospx
Ccospx
Dxcospx
Acoshpx
Ccoshpx
Dxcoshpx
Rule
Multiply the standard Particular Integral by x and then use
dy
d2 y
and
the product rule to find
dx
dx 2
1st order equations which can be solved using the 2nd order method
dy
b
 cy  f(x)
dx
d2 y
0
can be treated as a 2nd order differential equation with
dx 2
The following method only works if c is a number (not a P(x))
The solution will still consist of two parts:
y = complementary function + particular integral
dy
 3y  2e x
dx
Auxiliary equation
l–3=0
as
l2  0
Complementary function solution
y = Ae3x
Particular integral is of the form
y = Cex
l=3
dy
 3y  2e x
dx
P.I.
y = Cex
dy
 Ce x
dx
Substituting into the differential equation gives:
Cex – 3Cex = 2ex
–2C = 2
C = –1
The complete general solution is given by:
y = Complementary function + Particular integral
y = Ae3x – 1ex
dy
 3y  2e x
dx
Solution y = Ae3x – 1ex
Compare this with the method using an integrating factor
dy
 3y  2e x
dx
I.F.  e 
3dx
 e 3x
y  I.F   R.H.S  I.Fdx
y  e-3x   2ex  e3x dx
  2e2x dx
 e2x  C
Multiply through by e3x
y = Ce3x – 1ex
As was obtained above

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