Ordinary Differential Equations
Definition 1.1 : Differential Equation
An equation containing the derivatives of one or more
dependent variables, with respect to one or more independent
variables, is said to be a differential equation ( D E ).
Differential equations are classified by type, order and linearity.
Classification by Type :
If an equation contains only ordinary derivatives of one or more
dependent variables with respect to a single independent variable,
it is said to be an ordinary differential equation ( O D E ). For
example,
𝑑𝑦 𝑥 𝑑2 𝑦 𝑑𝑦
+ 10𝑦 = 𝑒 & 2 − + 6𝑦 = 0
𝑑𝑥 𝑑𝑥 𝑑𝑥
Are ordinary differential equations. An equation involving the
partial derivatives of one or more dependent variables with of two
or more independent variables is called a partial differential
equation ( P D E ). For example,
𝜕𝑢 𝜕𝑣 𝜕2 𝑢 𝜕2 𝑢 𝜕𝑢
=− & = −2
𝜕𝑦 𝜕𝑥 𝜕𝑥 2 𝜕𝑡 2 𝜕𝑡
Are partial differential equations.
Classification by Order :
The order of a differential equation ( O D E ore P D E) is the order
of the highest derivative in the equation. For example
𝑑2 𝑦 𝑑𝑦 3
𝟏. + 5 ( ) − 4𝑦 = 𝑒 𝑥
𝑑𝑥 2 𝑑𝑥
1
is a second-order ordinary differential equation.
𝟐. (𝑦 − 𝑥 )𝑑𝑥 + 4𝑥𝑑𝑦 = 0
Can be put into the form
𝑑𝑦
4𝑥 +𝑦 =𝑥
𝑑𝑥
is a first-order ordinary differential equation.
Classification as Linear or Nonlinear :
A differential equation 𝑦 (𝑛) = 𝑓(𝑥, 𝑦, 𝑦 ′ , … , 𝑦 (𝑛−1) ) is said to be
linear when 𝑓 is a linear function of 𝑦 ′ , … , 𝑦 (𝑛−1) . This means that
an equation is linear if it can be written in the form
𝑑𝑛 𝑦 𝑑𝑛−1 𝑦 𝑑𝑦
𝑎𝑛 (𝑥 ) ( )
𝑛 + 𝑎𝑛−1 𝑥
( )
𝑛−1 + ⋯ + 𝑎1 𝑥 + 𝑎0 (𝑥 )𝑦 = 𝑔(𝑥)
𝑑𝑥 𝑑𝑥 𝑑𝑥
The properties of linear differential equations:
i. The dependent variable 𝑦 and all derivatives are of the first
degree.
ii. Each coefficient depends only on the independent variable
𝑥.
Functions of y such as sin y or functions of derivatives of y such as
′
𝑒 𝑦 cannot appear in linear equation. A differential equation that
is not linear is said to be nonlinear.
The equations :
i. (𝑦 − 𝑥 )𝑑𝑥 + 4𝑥𝑑𝑦 = 0
′′ ′ 𝑑2 𝑦 𝑑𝑦
ii. 𝑦 − 2𝑦 + 𝑦 = 0, 2 −2 +𝑦 =0
𝑑𝑥 𝑑𝑥
2
𝑑3 𝑦 𝑑𝑦
iii. 𝑥 3 −4 + 6𝑦 = 𝑒 𝑥
𝑑𝑥 3 𝑑𝑥
Are linear first, second and third – order ordinary differential
equations, respectively.
The equations:
i. (1 + 𝑦)𝑦 ′ + 2𝑦 = 𝑒 𝑥
𝑑2 𝑦
ii. + 𝑠𝑖𝑛𝑦 = 0
𝑑𝑥 2
𝑑4 𝑦
iii. + 𝑦2 = 0
𝑑𝑥 4
Are nonlinear first, second and four- order differential equations,
respectively.
Definition 1.2 : Solutions of Differential Equation
Any function 𝜙 defined on some interval 𝐼 , which when
substituted into a differential reduces the equation to an identity,
is said to be a solution of the equation on the interval.
We say that 𝑦 = 𝜙(𝑥) satisfies the differential equation. The
interval 𝐼 could be an open interval (𝑎, 𝑏), a closed interval [𝑎, 𝑏] ,
an infinite interval (𝑎, ∞), and so on
Example 1 :
𝑥4 𝑑𝑦
Verify that 𝑦 = is a solution of the nonlinear equation =
16 𝑑𝑥
1
𝑥 𝑦 on the interval (−∞, ∞) .
2
Solution:
1
𝑑𝑦
Write the differential equation as −𝑥𝑦 =0
2
𝑑𝑥
3
1
1
𝑑𝑦 𝑥3 𝑥4 2 𝑥2
Using = 𝑎𝑛𝑑 𝑦 = ( ) =2
𝑑𝑥 4 16 4
1
𝑑𝑦 𝑥3 𝑥2 𝑥3 𝑥3
We see that − 𝑥 𝑦2 = −𝑥( ) = − =0
𝑑𝑥 4 4 4 4
Example 2 :
The function 𝑦 = 𝑥𝑒 𝑥 is a solution of the linear equation
𝑦 ′′ − 2𝑦 ′ + 𝑦 = 0
On the interval (−∞, ∞)
Solution:
We compute
𝑦 ′ = 𝑥𝑒 𝑥 + 𝑒 𝑥 & 𝑦 ′′ = 𝑥𝑒 𝑥 + 2𝑒 𝑥
Observe that
𝑦 ′′ − 2𝑦 ′ + 𝑦 = (𝑥𝑒 𝑥 + 2𝑒 𝑥 ) − 2(𝑥𝑒 𝑥 + 𝑒 𝑥 ) + 𝑥𝑒 𝑥 = 0
For every real number.
Example 3 : Verification of an implicit solution
The relation 𝑥 2 + 𝑦 2 − 4 = 0 is an implicit solution of differential
𝑑𝑦 𝑥
equation =−
𝑑𝑥 𝑦
On interval −2 < 𝑥 < 2
Solution:
By implicit differentiation we obtain
𝑑𝑦 2𝑥 2𝑦 𝑑𝑦 𝑑𝑦 𝑥
2𝑥 + 2𝑦 =0 ⟹ + =0 ⟹ =−
𝑑𝑥 2𝑦 2𝑦 𝑑𝑥 𝑑𝑥 𝑦
4
Example 4:
Verify that 𝑦 = 2 sin 𝑥 + 3 cos 𝑥 is a solution of the differential
𝑑2 𝑦
equation + 𝑦 = 0.
𝑑𝑥 2
Solution:
𝑑𝑦 𝑑2 𝑦
= 2 cos 𝑥 − 3 sin 𝑥 & = −2 sin 𝑥 − 3 cos 𝑥
𝑑𝑥 𝑑𝑥 2
𝑑2 𝑦
We see that + 𝑦 = (−2 sin 𝑥 − 3 cos 𝑥 ) + (2 sin 𝑥 +
𝑑𝑥 2
3 cos 𝑥 ) = 0
Example 5 : Using different symbols
The function 𝑥 = 𝑐1 cos 4𝑡 & 𝑥 = 𝑐2 sin 4𝑡 where 𝑐1 & 𝑐2 are
arbitrary constants, are solutions of the differential equation
𝑥 ′′ + 16𝑥 = 0
Solution:
For 𝑥 = 𝑐1 cos 4𝑡 the first two derivatives with respect to t are
𝑥 ′ = −4𝑐1 sin 4𝑡 & 𝑥′′ = −16𝑐1 cos 4𝑡
Substituting 𝑥 ′′ & 𝑥 then gives
𝑥 ′′ + 16𝑥 = −16𝑐1 cos 4𝑡 + 16(𝑐1 cos 4𝑡) = 0
In like manner
For 𝑥 = 𝑐2 sin 4𝑡 we have 𝑥 ′ = 4𝑐2 cos 4𝑡 & 𝑥 ′′ =
−16𝑐2 sin 4𝑡
5
Substituting 𝑥 ′′ & 𝑥 then gives
𝑥 ′′ + 16𝑥 = −16𝑐2 sin 4𝑡 + 16(𝑐2 sin 4𝑡) = 0
Finally two parameter family 𝑥 = 𝑐1 cos 4𝑡 + 𝑐2 sin 4𝑡 is a
solution of the given equation
Examples Exercises 1.1
State whether the given differential equation is linear or
nonlinear. Give the order of each equation.
𝟏. (1 − 𝑥 )𝑦 ′′ − 4𝑥𝑦 ′ + 5𝑦 = 𝑐𝑜𝑠𝑥
The equation is linear and second order.
Verify that the indicated function is a solution of the given
differential equation.
𝑥
𝟏𝟏. 2𝑦 ′ + 𝑦 = 0 ; 𝑦 = 𝑒 −2
′ 1 −𝑥
We compute 𝑦 = − 𝑒 2
2
Observe that
′ 1 −𝑥 𝑥
−2
2𝑦 + 𝑦 = 2 (− 𝑒 2 ) +𝑒 =0
2
𝑑𝑦
𝟏𝟑. − 2𝑦 = 𝑒 3𝑥 ; 𝑦 = 𝑒 3𝑥 + 10𝑒 2𝑥
𝑑𝑥
We compute 𝑦 ′ = 3𝑒 3𝑥 + 20𝑒 2𝑥
Observe that
6
𝑑𝑦
− 2𝑦 = 3𝑒 3𝑥 + 20𝑒 2𝑥 − 2(𝑒 3𝑥 + 10𝑒 2𝑥 ) = 𝑒 3𝑥
𝑑𝑥
𝟏𝟓. 𝑦 ′ = 25 + 𝑦 2 ; 𝑦 = 5 tan 5𝑥
We write the differential equation as 𝑦 ′ − 25 − 𝑦 2 = 0
We compute 𝑦 ′ = 25 𝑠𝑒𝑐 2 5𝑥
Observe that
𝑦 ′ − 25 − 𝑦 2 = 25 𝑠𝑒𝑐 2 5𝑥 − 25 − 25 tan2 5𝑥 = 25( 𝑠𝑒𝑐 2 5𝑥 −
tan2 5𝑥 − 1)
1 1
𝟏𝟕. 𝑦 ′ + 𝑦 = sin 𝑥 ; 𝑦 = sin 𝑥 − cos 𝑥 + 10𝑒 −𝑥
2 2
We write the differential equation as 𝑦 ′ + 𝑦 − sin 𝑥 = 0
1 1
We compute 𝑦 ′ = cos 𝑥 + sin 𝑥 − 10𝑒 −𝑥
2 2
Observe that
1 1 1 1
𝑦 ′ + 𝑦 − sin 𝑥 = ( cos 𝑥 + sin 𝑥 − 10𝑒 −𝑥 ) + ( sin 𝑥 − cos 𝑥 +
2 2 2 2
10𝑒 −𝑥 ) − 𝑠𝑖𝑛𝑥 = 0
