4.2 Ordinary Differential Equation (Homogeneous Equation)

2 = + 2 = +

Separating variables : = Write as a subject :
+

2 = +
2 1 = 2
+ =

MOOC MAT438/ UiTM

2 = + We can solve it first we
using separation create a new
of variable
variable

can also be written as

And, using product rule,

= ′ + ′ = =
′ =

′ =

we can solve the Differential Equation using
and

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ODEs must have the same degree.

Write as a subject.



Perform substituting using and

Solve the resulting differential equation using separation of variables
and

Solve the original differential equation in terms of x and y.

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MOOC MAT438/ UiTM

/ MAT235/ Jun 2019/ Q4c/ 8 marks

Find the general solution of the homogeneous differential equations :
2 + 2 + 1 1
= Same degree Homogeneous
2
(degree 2) equation

Step 1 Step 3

2 + 2 + 2 2 + 1 +
 + =
= 2 2

Step 2 + = 2 + 1 +


2 + 2 + Solve separable equation,
= 2
Step 4

2 2 + 2 + 2
+ =
2 = 2 + 1 + −

MOOC MAT438/ UiTM

/ MAT235/ Jun 2019/ Q4c/ 8 marks

Find the general solution of the homogeneous differential equations :

2 + 2 +
=
2

Step 4 Step 5

= 2 + 1 11
2 + 1 =
1
න 1 + 2 = −1 −1 = +

2 + 1 no need to find the
= value of c, since the
initial condition is not
Step 6 given.

11 the general solution is, Stop here!
2 + 1 =

−1 = + #


MOOC MAT438/ UiTM

MOOC MAT438/ UiTM

Solve the homogeneous first order differential equation given below.
4 1 1 = 2 − 2 Same degree Homogeneous
(degree 2)
equation

Step 1 Write as a subject and name as equation  Step 3 factorize and simplify


2 − 2  2 2 − 1 if we have a
= 4 + = 4 2 homogeneous equation
with degree 2, we will
succeed in factoring 2

Step 2 substitute and = into  2 − 1
= + + = 4

2 − 2 Solve separable equation,
= 4 Step 4 separating variables v and x

2 2 − 2 2 − 1
+ = 4 2 = 4 −

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Solve the homogeneous first order differential equation given below.
4 = 2 − 2

Step 5

Step 4

4 1 4
−3 2 − 1 = −1 න 3 2 + 1 = −4 න 3 2 + 1
2 − 1
= 4 − 4 1 = 3 2 + 1
= −4 න ∙ 6
2 − 1 − −1 න 3 2 + 1 = න = 6
= 41
4 1 = − 6 න
−4 න 3 2 + 1 = න = 6
21
2 − 1 − 4 2 = − 3 න
=
4 2 3 2
3
−3 2 − 1 − + 1 = + 2
= 4 = − 3 +

no need to find the Step 6 = − 2 3 2 + 1 +
value of c, since the 3
4 1
−3 2 − 1 = initial condition is the general solution is, 2 3 2 Stop here!
not given. − 3 + 1 = + #

MOOC MAT438/ UiTM

MOOC MAT438/ UiTM

Solve the initial value problem Homogeneous
3 2 + 9 1 1+ 5 2 − 6 2 + 4 1 1 = 0 ; 1 = 0 Same degree equation

(degree 2)

Step 1 Write as a subject and name as equation  Step 3 factorize and simplify

if we have a

3 2 + 9 + 5 2 = 6 2 + 4 2 3 + 9 + 5 2 homogeneous equation
+ = 2 6 + 4 with degree 2, we will
succeed in factoring 2

3 2 + 9 + 5 2  3 + 9 + 5 2
= 6 2 + 4 + = 6 + 4

Step 2 substitute and = into 
= +
Solve separable equation,
3 2 + 9 + 5 2 Step 4 separating variables v and x
= 6 2 + 4
3 + 9 + 5 2
3 2 + 9 2 + 5 2 2 = 6 + 4 −
+ =
6 2 + 4 2

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Solve the initial value problem
3 2 + 9 + 5 2 − 6 2 + 4 = 0 ; 1 = 0.

Step 4 separating variables v and x Step 5 integrating both sides

3 + 9 + 5 2 4 + 6 1 LHS : integration by u-substitution
= 6 + 4 −
2 + 3 + 3 = 4 + 6 4 + 6
න 2 + 3 + 3 = න ∙ 2 + 3

3 + 9 + 5 2 − 2 2 + 3 + 3 = + = 2 + 3 + 3 = න2 2 + 3
∙ 2 + 3
= = 2 + 3
6 + 4 Step 6 replace v = 2
= න
3 + 9 + 5 2 − 6 − 4 2 = 2 + 3
= 1
6 + 4 if we choose the = 2 න
න = +
the general solution is, correct u, we will
3 + 3 + 2 successfully write all = 2 +
= 6 + 4
terms in u and du

2 2+3 +3 = + = 2 2 + 3 + 3 +


4 + 6 1
2 + 3 + 3 =

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Solve the initial value problem
3 2 + 9 + 5 2 − 6 2 + 4 = 0 ; 1 = 0. To find the value of C

(after integrate)
= 1 = 0

the general solution is, Thus, the particular solution is,

2 2+3 +3 = + ❷ 2+3

2 +3 = + 9 #

Step 7

Substitute the initial condition = 1, = 0 into ❷

0

2 3 = 1 +

= 2 3 log = log

= 9 Into ❷

MOOC MAT438/ UiTM

MOOC MAT438/ UiTM

Solve the homogeneous first order differential equation , 3 − 2 3 Same degree
= − 1 2 (degree 3)

Step 1 Write as a subject and name as equation  Step 3 factorize and simplify Homogeneous
equation

3 − 2 3  3 1 − 2 3 if we have a
= − 2 + = − 2 3 homogeneous equation
with degree 3, we will
1 − 2 3 succeed in factoring 3
+ = − 2
Step 2 substitute and = into 
= +

3 − 2 3 Solve separable equation,
= − 2 Step 4 separating variables v and x
3 − 2 3 3
+ = − 2 3 1 − 2 3
= − 2 −

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Solve the homogeneous first order differential equation , 3 − 2 3 Same degree
= − 1 2 (degree 3)

1 − 2 3 − 3 − 1 Homogeneous
= − 2 − = − 2 equation

1 − 2 3 − 3 − 1
= = 2
− 2 2 1
3 − 1 =
1 − 2 3 + 3
= Step 5 integrating both sides
− 2
2 1
1 − 3 න 3 − 1 = න
= − 2

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Solve the homogeneous first order differential equation , 3 − 2 3
= − 2

the degree of the top

is less than the degree

degree 2 of the bottom and the
difference is 1
2 1
න 3 − 1 = න 3 − 1 = 3 2 Can use u-subs and
u=denominator. The answer will
degree 3
2 1 give ‘ln’ after integration.
න 3 − 1 = න
Numerator is a 2 2
derivative of න 3 − 1 = න ∙ 3 2
denominator

1 3 − 1 = + the answer will give = 3 − 1 11
3 ln|denominator| after = 3 න
= 3 2
integration 1
= 3 +
3 − 1 = 3 +
= 3 2

Step 6 replace v = 1
= 3
the general solution is, 3 − 1 +

3−1 = 3 + #


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Solve the homogeneous first order differential equation , 3 − 2 3
= − 2

Answer : 3−1 = 3 + 3 2 − 3 2 3 3


3−1 = 3 + 3 − 3 − =

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Or… wrong, maybe ??? 3 cN2 o o 3m ws −p u,a−bw r jee33ec tw to2a,nt=sthoet 3t ogh+iwavter3 inwteequ c e a snatsioan

3 hmmm
3 − 1 = 3 +

3 − 3 3 2 − 2 3+3
3 = 3 + = − =

3 − 3 − 3 = 3 + 3 − 3

Differentiate implicitly wrt x 3 2 − 2 6
=
1 3 2 3 2 1 3 2 1
3 − 3 ∙ − − 3 ∙ = 3 ∙ + 0 3 − 3

At this point, we done
differentiate!

MOOC MAT438/ UiTM

Solve the homogeneous first order differential equation , 3 − 2 3
= − 2

Answer : 3−1 = 3 + 2 3 − 2 3 + 3


2 2 2 =

3 − 6
=
3 − 3 2 = 2 3 − 3

26 3 − 3
2 − 2 = 3

2 3 − 3 At this point, we
= 2
2 3 − 3 success write as a

2 − 2 =
subject

2 3 2 3 − 3 − 2 3 −1 1 1
− = 2 = −2 = −2
2 − 2 = 2

2 = 2 3 − 2 3 + 2 3 − 2 3 3 − 2 3
= − 2 = − 2

MOOC MAT438/ UiTM

MOOC MAT438/ UiTM