MATH 251: DIFFERENTIAL EQUATIONS
M. A. BOATENG, PhD., MIMA
March 3, 2022
TABLE OF CONTENT
Definition
Classification
Solution to a Differential Equation
Methods of Solving Differential Equations
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DEFINITION
A differential equation is an equation with an unknown
function and one or more of its derivative.
A differential equation is an equation relating some
function f to one or more of its derivatives.
Example:
d 2f
df
+ 2 + f 2 x = sinx
2
dx
dx
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CLASSIFICATION
Differential equation can be classified by the following:
Type: The equation of derivatives are taken with
respect to one or several variables. Example
dx
= yx (ODE )
dy
(2)
δy δy
+
= 0(PDE )
(3)
δt
δx
Order: The order of a DE is the order of the highest
derivative in that particular equation.
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Examples.
dy 2 d 2 y
) + 2 = sinx (order 2)
dx
dx
3
d x
+ x ll + tx 2 = yx (order 3)
3
dy
(
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Linearity: A DE is said to be linear if the following
conditions are satisfied;
the dependent variable(s) and their derivatives appear
linear or in the power of one.
There is no product of the dependent variable and any
of its derivatives.
There is no transcendental function (Trignometric,
Logarithmic or exponential) of the dependent
variable).
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Examples
d 4x
= 0(linear )
dt 4
x ll + tx 2 = t(non − linear )
(6)
(7)
Homogeneity: if the equation is not equal to a
function of the independent variable.
Example
dy 2 d 2 y
+ 2 = 0(homogeneous)
dx
dx
d 3x
+ x ll = sinx (non − homogeneous)
3
dy
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Degree: The degree is the exponent of the highest
derivative.
Example
dy
dx
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!2
+
d 2y
= sinx (degree1)
dx 2
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SOLUTION TO A DIFFERENTIAL EQUATION
A function that satisfies the given differential equation is
called its solution.
The solution that contains as many arbitrary
constants as the order of the differential equation is
called a general solution.
The solution free from arbitrary constants is called a
particular solution.
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Examples
−3
Show that y (x ) = x 2 is a solution to the differential
equation
4x 2 y ′′ + 12xy ′ + 3y = 0 for x > 0
solution
−5
2
y ′ = −3
2 x
−7
2
y ′′ = 15
4x
substituting this into the question yields
−7
−3
−3 −5
2 ) + 12x (
2 ) + 3(x 2 ) = 0
4x 2 ( 15
x
x
4
2
0=0
EXAMPLE
Find the DE whose general equation is
y = c1 e −2x + c2 e 3x
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SOLUTION TO A DIFFERENTIAL EQUATION
There are two forms of solution to a differential equation.
They are:
General Solution
Particular Solution
General Solution
A general solution is a function that gives all the possible
solutions to a differential equation.
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Example: Solve the ODE
dy
dx
= −4
Solution
R
dy = −4dx
R
y = −4x + c where c is an arbitrary constant.
This type of solution is a general solution.
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Particular Solution
A Particular Solution is a solution of a differential
equation taken from the General Solution by allocating
specific values to the random constants.
Example: Solve y ′ = −4, y (0) = 1
from the general solution we have
y = −4x + c
substituting the condition given yields
1 = 4(0) + c
c=1
Hence the particular solution becomes
y = −4x + 1
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SUBSIDIARY CONDITIONS
SUBSIDIARY CONDITION
A subsidiary condition helps to find a particular solution
to a differential equation.
There are two types of subsidiary conditions.
Initial conditions
Boundary conditions
Example:
1
y ′ = −4, y (0) = 1 (Initial condition)
′′
2
y = −4, y (0) = 1, y ′ (0) = 2 (Initial condition)
3
y ′′ + 2y ′ = e 2 , y (0) = 1, y ′ (1) = 1 (boundary (0,1)
condition)
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NB: Both boundary and initial conditions gives a
particular solution.
EXAMPLE
Given that y = c1 e −x + c2 e 3x is a general solution of the
D.E y ′′ − 2y ′ − 3y = 0, y (0) = 3, y ′ (0) = 4. Find the
particular solution.
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SOLUTION
y = c1 e −x + c2 e 3x
3 = c1 + c2
y ′ = −c1 e −x + 3c2 e 3x
4 = −c1 + 3c2
Solving simultaneously yields
c1 = 54 and c2 = 74
Hence the particular solution becomes
y = 54 e −x + 74 e 3x
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SOLUTIONS TO FIRST ORDER ODEs
First order ODEs are in the form F (x , y , y ′ ) = 0 There
two forms ODEs. They are
Standard Form
y ′ = f (x , y )
dy
dx = f (x , y )
Differential form.
M(x,y)dx+N(x,y)dy=0
(x + y )dx + y 2 dy = 0
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METHODS OF SOLVING ODEs
Separation of variables
Homogeneous Method
Integrating factor
Exactness
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METHOD OF SEPARATION OF VARIATION
M(y ) dy
dx = N(x )
M(x , y )dy = N(x , y )dx
let M(x,y)=m(x).p(y)
N(x,y)=n(x).q(y)
m(x).p(y)dx + n(x).q(y)dy=0
m(x ) p(y )
n(x ) q(y )
n(x ) . p(y ) dx + n(x ) . p(y ) dy = 0
=⇒
R m(x )
n(x ) dx
+
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R q(y )
p(y ) dy
=c
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Example
Solve the ODE
2
(x + 1) dy
dx = x (y + 1)
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Solution
(x + 1)dy = x (y 2 + 1)dx
(x + 1)dy = (xy 2 + xdx )
(x + 1)dy − (xy 2 + xdx ) = 0
x
1
x +1 dx − y 2 +1 dy = 0
R x
R 1
dx
−
x +1
y 2 +1 dy = C
let u = x + 1
x =u−1
R u+1
R 1
u du − y 2 +1 dy = C
R
R
R
udu u1 du − y 21+1 dy = C
(x + 1) − ln|x + 1| − tan−1 y = C
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TRIAL
Solve the following ODEs
dy
x2
1
=
dx
y (1+x 3 )
2
3
(1 + x 2 ) − x (1 + 2y )y ′ = 0
2
−4
y ′ = 3x 2y+4x
−4 , y (1) = 3
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HOMOGENEOUS FUNCTION
Definition
A function f(x,y) is homogeneous if f (λx , λy )=λn f (x , y ),
n = −∞, ., ., .∞
Example: Check if f (x , y ) = x 4 − x 3 y is homogeneous.
solution
f (λx , λy ) = (λx )4 − (λx )3 (λy )
f (λx , λy ) = λ4 x 4 − λ3 x 3 λy
f (λx , λy ) = λ4 x 4 − λ4 x 3 y
=⇒ f (λx , λy ) = λ4 (x 4 − x 3 y )
Hence homogeneous.
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Trials
Check for homogeneity in the following function
+y
f (x , y ) = 2x
x 2y 2
y
f (x , y ) = e x + tan( yx )
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HOMOGENEOUS METHOD
In using the homogeneous method, we follow the
following rules:
Check for homogeneity of the function.
Divide through the function by the highest term of the
independent variable.
Let v = yx
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EXAMPLES
Determine whether the differential equation is
homogeneous, hence solve it.
(x 2 + y 2 )dx − 2xydy = 0
SOLUTION
(x 2 +y 2 )
dy
=
dx
2xy
From the funciton on the right hand side, we could see
that, each term is of the power two(2). Hence, the
function is homogeneous.
Let v = yx
dy
dv
dx = v + x dx
dy = vdx + xdv
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(x 2 + (xv )2 )dx − 2x (xv )(xdv + vdx ) = 0
x 2 dx + x 2 v 2 dx − 2x 3 vdv − 2x 2 v 2 dx = 0
x 2 dx
2x 3 v
x 2v 2
−
−
2
2
x
x
x2 = 0
dx − v 2 − 2xvdv = 0
(1−v 2 )
2xv
2 dx − x dv = 0
1−v
R 1
R 2v
x dx − 1−v 2 dv = 0
ln |x | = ln |C | − ln |1 − v 2 |
C
ln |x | = ln 1−v
2
C
x = 1−v 2
but v = yx
x = 1−(Cy )2
x
x=
y=
x 2C
x 2 −y 2
√
x2 −
Cx
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TRIAL
1
2
3
4
2
dy
dx
2
+y
= x xy
2xdy = (x + y )dx
dy
y 2 −x 3
dx = yx
dy
dx
=
2x +y
x 2y 2
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HOMOGENEOUS LINEAR COEFFICIENT FUNCTIONS
Given an ODE in the form
(a1 x + b1 y + c1 )dx + (a2 x + b2 y + c2 )dy = 0,
we say this type of ODE is almost homogeneous. There
are two ways of solving the problem.
Case 1:
When aa21 ̸= bb12
Case 2:
When aa21 = bb12
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CASE 1
Given an ODE in the form
(a1 x + b1 y + c1 )dx + (a2 x + b2 y + c2 )dy = 0
When aa21 ̸= bb12
Set x = X + h, y = Y + k
where (a1 X + b1 Y )dx + (a2 X + b2 Y )dy = 0
(a1 h + b1 k + c1 ) = 0
(a2 h + b2 k + c2 ) = 0
Then we solve for h and k.
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Example
Solve the first order ODE
(x − 2y + 1)dx + (4x − 3y − 6)dy = 0
Solution
a2
b2
4
3
a1 = 1 ̸= b1 = 2
Set x = X + h, y = Y + k
h − 2k + 1 = 0
4h − 3k − 6 = 0
Solving this simultaneous h = 3, k = 2
(X − 2Y )dX + (4X − 3Y )dY = 0
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dY
dX
X −2Y
= − 4X
−3Y
=
1−2 YX
3 YX −4
1−2v
vdX + xdv = 3v
−4 dX
2
3v −2v −1
dX = −Xdv
3v
R −4
R
1
−4
− X dX = 3v 23v−2v
−1
− ln |X | = 15
ln
|3v
+ 1| − 14 ln |v − 1| + ln |c|
4
−4 ln |X | = 15 ln |3v + 1| − ln|v − 1| + 4 ln |c|
15
ln |X −4 c 4 | = ln | (3vv+1)
−1 |
X −4 c 4 =
X −4 c 4 =
X −5 c 4 =
(3v +1)15
v −1
(( YX )+1)15
( YX )−1
( 3YX+x )15 /(Y
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15
)
X 10 c 4 = (3YY+X
−X
X 10 c 4 = (3Y + X )15
(x − 3)10 c 4 Y − X = (3y − 6 + x − 3 − 9)15
(x − 3)10 (y + x − 5)c 4 = (3y + x − 9)15
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Trials
1
2
(2x + y − 2)dx + (−x + 2y + 1)dy = 0
(x − 3y )dx + (x + y − 4)dy = 0
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case 2
Given an ODE in the form
(a1 x + b1 y + c1 )dx + (a2 x + b2 y + c2 )dy = 0
When aa21 = bb12 = K
Set z = a1 x + b1 y
Example:
Solve the ODE (x + 2y + 3)dx + (2x + 4y − 1)dy = 0
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Solution
a2
b2
2
4
a1 = 1 = b1 = 2 = 2
let z = x + 2y
dy
dz
dx = 1 + 2 dx
dz−dx
= dy
2
(z + 3)dx + (2z − 1)( dz−dx
2 ) =0
2(z + 3)dx + (2z − 1)(dz − dx ) = 0
(2z + 6)dx + 2zdz − 2zdx − dz + dx = 0
R
R
7dx + (2z − 1)dz = 0
7x + z 2 − z = c
7x + (x + 2y )2 − (x + 2y ) = c
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Trial
1
2
(x + 2y + 1)dx + (2x + 4y + 3)dy = 0
(x − y )dx + (x − y + 2)dy = 0
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INTEGRATING FACTOR
In order to solve a linear first order differential equation
we MUST start with the differential equation in the form
dy
dt + p(t)y = g(t) where both p(t) and g(t) are
continuous functions.
We find the solution to such a problem by multiplying
a
R
function called the integrating factor µ(t) = e p(t)dt
through the DE. i.e. µ(t) dy
dt + µ(t)p(t)y = µ(t)g(t).
This reduces to dtd µ(t)y = µ(t)g(t) (A derivative of the
product of the integrating factor and the dependent
variable).
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Example
Solve the differential equation
dy
dt − 2y = 4 − t
SolutionR
µ(t) = e −2dt .
µ(t) = e −2t
−2t
e −2t dy
− te −2t
dt = 4e
e −2t dy = (4e −2t − te −2t )dt
R −2t
R
e dy = (4e −2t − te −2t )dt
−2t
− 14 e −2t ) + C
e −2t y = −2e −2t − ( −t
t e
−2t
e −2t y = −2e −2t + −t
− 14 e −2t + C
t e
y = 2t − 47 + ce 2t
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Exercise
Solve the following IVP differential equations
dv
dt = 9.8 − 0.196, y (0) = 48
ty ′ + 2y = t 2 − t + 1, y (1) = 21
ty ′ − 2y = t 5 sin(2t) − t 3 + 4t 4 , y (π) = 23 π 4
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