4.5.1 Second Order Differential Equations (Homogeneous Equation)

MOOC MAT438/ UiTM

At the end of this session, the students should be able to :

Solve and for Second
order ODE

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Where a, b, c are constants, and a  0

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m2 m 1

Distinct reaml 1roots m( 0 1 ≠ 2)

Repeated real roots ( 1 = 2) Just write

m only

iWhat is ??

Complex roots ( = ± )

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What is ?? is an imaginary −

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=

−1 = Just leave
as ‘surd’

−2 = 2 −1 = 2 −1 = 2

−3 = 3 −1 = 3 −1 = 3 Simplify
−4 = 4 −1 = 4 −1 = 2 the perfect

square

Solve the second order homogeneous equation

′′− ′ − 2 = 0, ℎ = 0, = 3, ′ = 0 the initial conditions is given !

m1 m0

m2 m 1

From ′′ − ′ − 2 = 0 Step 3 Write the general solution (based on the
types of roots)
Characteristic Equation
Step 1 2 − − 2 = 0 general solution,

= 1 1 + 2 2

Step 2 Solve Quadratic Equation = 1 − + 2 2 (this is the final answer, if the initial

+ 1 − 2 = 0 If the solution is complex conditions is not given)
1 = −1, 2 = 2 roots, never use calculator.
if the initial conditions is given,
use quadratic formula
Step 4 Find y’ (differentiate the general solution)
1 ≠ 2 Distinct real roots ′ = − 1 − + 2 2 2

MOOC MAT438/ UiTM = 1, = −1, = −2

− = −1 2 − 4 1 −2
=1+8
=9

− > Distinct real roots

Solve the second order homogeneous equation
′′− ′ − 2 = 0, ℎ = 0, = 3, ′ = 0

= 1 − + 2 2 0 = − 1 + 2 2 
1 = 2 2

′ = − 1 − + 2 2 2 Step 7 Solve equation  and  simultaneously

Step 5 Substitute the initial condition = 0, (to find C1 and C2)

= 3 into (naming it as equation ) Substitute  into  : 3 = 2 12 + 2

3 = 1 0 1 2 0 1 3 = 3 2

+ 2 = 1 (into )

3 = 1 + 2 

from : 1 = 2 12 = 2

Step 6 Substitute the initial condition = 0,

′ = 0 into ′ (naming it as equation ) Step 8 Write the particular solution (substitute back
1
0 = − 1 0 2 2 0 1 the values of C1 and C2 into the general solution )
+
= 21 − + 12 2 #

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Solve the second order homogeneous equation
′′− ′ − 2 = 0, ℎ = 0, = 3, ′ = 0

is my answer ✓ = 2 − + 2 Substitute into the given question :
correct ?? ′′− ′ − 2 = 0
Or… wrong, ′ = −2 − + 2 2

maybe ??? hmmm ′′ = 2 − + 4 2

′′− ′ − 2 = 20 − + 4 2 − −2 − + 2 2 −2 2 − + 2

= 2 − 0 4 2 0

+

+2 − − 2 2

−4 − − 2 2

′′− ′ − 2 = 0 Shown !

MOOC MAT438/ UiTM

MOOC MAT438/ UiTM

Solve the second order homogeneous equation

′′−2 ′ + = 0 the initial conditions is not given !

m1 m0

m2 m 1

From ′′ − 2 ′ + = 0 Step 3 Write the general solution (based on the
types of roots)
Characteristic Equation
Step 1 2 − 2 + 1 = 0 general solution,

= 1 + 2

Step 2 Solve Quadratic Equation = 1 + 2 # (this is the final answer, because

If the solution is complex roots, the initial conditions is not given)

− 1 − 1 = 0 never use calculator. We need

to use quadratic formula !

= 1, = 1 = 1, = −2, = 1
1 = 2 Repeated roots
− = −2 2 − 4 1 1
=4−4
=0

− = Repeated roots

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Solve the second order homogeneous equation

′′+ = 0 the initial conditions is not given !

m1 m0

m2 m 1 Write the general solution (based on the
types of roots)
From ′′ + 0 ′ + = 0 Step 3

Step 1 Characteristic Equation general solution,
2 + 1 = 0 = 1 + 2

Step 2 Solve Quadratic Equation = 0 1 1 + 2 1

2 = −1 If the solution is complex roots, (this is the final answer,
= ± −1 never use calculator. We need to
= 1 + 2 # because the initial conditions
use quadratic formula ! is not given)

complex roots = ± = 0 ± 1 = 1, = 0, = 1 0 ± −
− = 0 2 − 4 1 1 = 2 1

= ± =0−4 0 ± −
= 0 = 1 = −4 = 2
− < complex roots
0 ± 2
Using quadratic formula, = 2

MOOC MAT438/ UiTM − ± − 02
= 2 = 2 ± 2
= 0 ± 1

Solve the second order homogeneous equation
′′+ = 0

is my answer ✓ = 1 + 2 Substitute into the given question :
correct ?? ′′+ = 0
Or… wrong, ′ = − 1 + 2

maybe ??? hmmm ′′ = − 1 − 2

′′+ = −0 1 − 2 + 1 + 2

′′+ = 0 Shown !

MOOC MAT438/ UiTM

MOOC MAT438/ UiTM