1. SECOND ORDER
LINEAR
DIFFERENTIAL
EQUATIONS

2. We shall consider solutions to equations of the form:
d2 y
d x2 dy
d x a
+ b
+ cy = f(x)
ay + by + cy = f(x)
or:
We shall see that the solution consists of two parts, the
complementary function (C.F.) which is the solution to
The equation when f(x) = 0, and the particular integral,
(P.I.) which relates to f(x) when f(x) ≠ 0.

3. The complementary function
If we guess that a solution of
ay + by + cy = 0
is y = eαx with α to be determined:
So: y = eαx
y =
αeαx
y =
α2eαx
Substitute these into the original equation:
Then: aα2eαx + bαeαx + ceαx = 0
eαx ( aα2 + bα + c ) = 0
Hence y = epx and y = eqx , where p and q are the roots
of aα2 + bα + c = 0 , are solutions.
i.e. The C.F. is y = Aepx + Beqx.

4. We must now consider the two special cases which arise when
aα2 + bα + c = 0 ( called the auxiliary equation ) has either
Two equal roots, or ii) Complex roots.
i) If p = q
By observing the sum and product of the roots:
= – 2p, and b a c
= p2
The differential equation can then be written:
y – 2py + p2y = 0 … (1)
Let y = epxg(x), where g(x) is some function of x.
y = e pxg (x) + g(x)pe px
y = e pxg (x) + 2pg (x)e px + p2g(x)e px
In (1):
e px[ g (x) + 2pg (x) + p2g(x) – 2pg (x) – 2p2g(x) + p2g(x) ] = 0
e pxg (x) = 0
g (x) = 0
g(x) = Ax + B
Hence the C.F. is:
y = epx (Ax + B )

5. ii) If p and q are complex, we can write: p = a + ib, q = a – ib.
The C.F. is y = Ae(a + ib)x + Be(a – ib)x
= Aeaxeibx + Beaxe–ibx
= eax( Aeibx + Be–ibx )
= eax [A(cosbx + isinbx) + B(cos(–bx) + isin(–bx))]
= eax [A(cosbx + isinbx) + B(cosbx – isinbx)]
= eax [Ccosbx + Dsinbx]
( where C and D are constants, which include complex values)

6. The solutions to the differential equation: d2 y d x2 dy d x a + b
+ cy = 0
depend on the roots of the auxiliary equation:
au2 + bu + c = 0
If b2 – 4ac > 0 and has roots p and q
y = Aepx + Beqx
If b2 – 4ac = 0 with repeated root p
y = epx (A + Bx)
If b2 – 4ac < 0 with complex roots p  iq
y = epx ( Acosqx + Bsinqx )

7. Obtain the general solutions of the following equations:
Example 1:
Obtain the general solutions of the following equations:
i) 4y – y – 5y = 0
ii) y – 4y + 13y = 0
i) The auxiliary equation is:
4u2 – u – 5 = 0
– 1
u = 5 4 ,
Factorising this:
( 4u – 5 ) ( u + 1 ) = 0
y = Ae + Be – x 5x 4
The general solution is:
ii) The auxiliary equation is:
u2 – 4u + 13 = 0
= 2 ± 3i
The general solution is:
y = e2x ( Acos3x + Bsin3x)

8. d2 y
d x2 dy
d x
– 6
+ 9y = 0
Example 2:
Solve the differential equation: dy
d x
= 7.
given that when x = 0, y = 1 and
The auxiliary equation is:
u2 – 6u + 9 = 0
( u – 3 ) ( u – 3 ) = 0
u = 3
( repeated root )
The general solution is:
y = e3x (Ax + B) …(1) dy
d x =
Differentiating (1):
e3xA +
(Ax + B)3e3x …(2)
Now, if x = 0, y = 1 in (1):
1 = 1 ( 0 + B )
so, B = 1 dy
d x
= 7 in (2):
Also,if x = 0,
7 = 1 ( A ) + ( ) 3
so, A = 4
The solution is:
y = e3x (4x + 1 )

9. The particular integral
when f(x) ≠ 0.
Now, returning to the equation:
d2 y
d x2 dy
d x a
+ b
+ cy = f(x)
To establish the particular integral (P.I.), certain expressions for f(x)
suggest standard trial solutions:
f(x)
Trial solution p
y = λ
px + q
y = λx + μ
psinx + qcosx
y = λsinx + μcosx
e kx
y = λe kx
The solution to the differential equation is then y = C.F + P.I.

10. Example 3: Find the general solution of the differential equation:
d2 y
d x2 dy
d x
– 2
– 3y = e 2x
The auxiliary equation is:
u2 – 2u – 3 = 0
( u – 3 )( u + 1) = 0
u = 3, – 1
The C.F. is:
y = Ae3x + Be – x dy
d x =
d2 y
d x2 =
For the P.I. try
y = λe2x
2λe2x
4λe2x
Substitute these into the original equation:
λ = 1 3 –
4λe2x – 2(2λe2x) – 3(λe2x) = e2x
y = Ae3x + Be – x 1 3
e2x –
The general solution is:

11. Obtain the general solutions of the following equation:
Trials that fail
Example 4:
Obtain the general solutions of the following equation:
y – 3y + 2y = e2x.
The auxiliary equation is:
u2 – 3u + 2 = 0
( u – 1 )( u – 2) = 0
u = 1, 2
The C.F. is:
y = Aex + Be 2 x
For the P.I. we would normally try y = λe2x, but this is already part of
the solution.
Try y = λ x e2x
y = λx2e2x + e2x λ
= λe2x ( 2x + 1 ) y
= λe2x 2 + ( 2x + 1 )2λe2x
= 4λe2x ( x + 1 )
Sub. into the original equation:
4λe2x ( x + 1 )
– 3 λ e2x ( 2x + 1 )
+ 2 λ x e2x
= e2x
λe2x = e2x
λ = 1
The general solution is:
y = Aex + Be 2 x
+ xe2x

12. The solution to the differential equation: d2 y d x2 dy d x a + b
Summary of key points:
The solution to the differential equation:
d2 y
d x2 dy
d x a
+ b
+ cy = f(x)
is y = C.F + P.I.
The C.F. is found from the auxiliary equation:
au2 + bu + c = 0
If b2 – 4ac > 0 and has roots p and q, y = Aepx + Beqx
If b2 – 4ac = 0 with repeated root p, y = epx (A + Bx)
If b2 – 4ac < 0 with complex roots p  iq, y = epx(Acosqx + Bsinqx)
To establish the P.I. the given
trial solutions are possible.
f(x)
Trial solution p
y = λ
px + q
y = λx + μ
psinx + qcosx
y = λsinx + μcosx
e kx
λe kx
If the trial solution is part of
the C.F. then try multiplying by x. ( If this also fails, try
multiplying by x2 ).
This PowerPoint produced by R.Collins ; Updated Feb.2010