th
14 February 2026
First Order Differential Equations 2
Peter J Omale
Contents
Linear differential equations.
Equations reducible to the linear equation (Bernoulli’s equation).
Exact differential equations.
Performance objectives
At the end of this programme, students should be able to:
Solve linear differential equations.
Reduce Bernoulli’s equation to linear form
Solve Bernoulli’s equation.
Show that a given differential equation is exact.
Solve exact differential equations.
Linear differential equations
A linear differential equation is any equation of the form
dy
+ Py = Q
dx
Where P and Q are functions of x or constants.
Example 1
dy 3x
Solve - 5y = e
dx
dy
Compare with + Py = Q
dx
3x
P = -5 Q=e
∫Pdx=∫-5dx = -∫5dx=-5x
Redistribution of this material by photocopying or any other means is prohibited matrojek
1
, th
14 February 2026
∫Pdx
Integrating factor IF necessary for solving this problem is given by e .
. ∫Pdx -5x
. . IF = e =e
The solution is
∫
y.IF = Q .IF. dx
-2x
e
= ∫e ∫
-5x 3x -5x -2x
y.e .e dx = e dx = +c
-2
-2x -2x
e e 1
=- + c = c- = c - 2x
2 2 2e
y 1
5x = c - 2x
e 2e
5x
Multiply through by e
3x
e 5x
y = ce -
2
Example
dy 4
Solve x - 2y = x
dx
dy 4
x - 2y = x
dx
Divide through by x
dy 2 3
- y=x
dx x
Compare with
dy
+ Py = Q
dx
2 3
P =- Q=x
x
2 1
∫Pdx = ∫- x dx = -2∫ x dx = -2lnx = lnx -2
∫pdx lnx
-2
-2 1
I.F= e =e =x = 2
x
Redistribution of this material by photocopying or any other means is prohibited matrojek
2
14 February 2026
First Order Differential Equations 2
Peter J Omale
Contents
Linear differential equations.
Equations reducible to the linear equation (Bernoulli’s equation).
Exact differential equations.
Performance objectives
At the end of this programme, students should be able to:
Solve linear differential equations.
Reduce Bernoulli’s equation to linear form
Solve Bernoulli’s equation.
Show that a given differential equation is exact.
Solve exact differential equations.
Linear differential equations
A linear differential equation is any equation of the form
dy
+ Py = Q
dx
Where P and Q are functions of x or constants.
Example 1
dy 3x
Solve - 5y = e
dx
dy
Compare with + Py = Q
dx
3x
P = -5 Q=e
∫Pdx=∫-5dx = -∫5dx=-5x
Redistribution of this material by photocopying or any other means is prohibited matrojek
1
, th
14 February 2026
∫Pdx
Integrating factor IF necessary for solving this problem is given by e .
. ∫Pdx -5x
. . IF = e =e
The solution is
∫
y.IF = Q .IF. dx
-2x
e
= ∫e ∫
-5x 3x -5x -2x
y.e .e dx = e dx = +c
-2
-2x -2x
e e 1
=- + c = c- = c - 2x
2 2 2e
y 1
5x = c - 2x
e 2e
5x
Multiply through by e
3x
e 5x
y = ce -
2
Example
dy 4
Solve x - 2y = x
dx
dy 4
x - 2y = x
dx
Divide through by x
dy 2 3
- y=x
dx x
Compare with
dy
+ Py = Q
dx
2 3
P =- Q=x
x
2 1
∫Pdx = ∫- x dx = -2∫ x dx = -2lnx = lnx -2
∫pdx lnx
-2
-2 1
I.F= e =e =x = 2
x
Redistribution of this material by photocopying or any other means is prohibited matrojek
2