BUENAVISTA COMMUNITY COLLEGE
Cangawa, Buenavista, Bohol
Telefax: (038)5139169/Tel.: 513-9179
Differential Equations of First Order and First Degree
I. Overview:
A differential equation of first order and first degree is in the form of
M(x,y) dx + N(x,y) dy = 0
II. Specific Objectives: After studying the material presented in the text(s), lecture,
computer tutorials, and other resources, the student should be able to complete all
behavioral/learning objectives listed below with a minimum competency of 70%:
a) Identify homogeneous equations, homogeneous equations with constant
coefficients, and exact and linear differential equations.
b) Solve ordinary differential equations and systems of equations using:
b.1) direct integration
b.2) separation of variables
b.3) homogeneous equations
c) Determine general solutions to differential equations
III. Learning Session:
Any differential equation of the first order and first degree can be written in the form
dy
= F (x,y) or M(x,y) dx + N(x,y) dy = 0
y
where M and N are functions of x and y.
Illustration. The differential equation
dy x−3 y
y
= 2 y −x
can also be written as
(x - 3y)dx + (x - 2y)dy = 0
Methods of solving differential equations of the first order and first degree
A. Separation of variables. If an equation
M(x, y) dx + N(x, y) dy = 0
can be brought into the form by algebraic process
M1(x) dx + N1(y) dy = 0,
where M1 and N1 are functions of one variable as indicated, we say that the
variables have been separated. The general solution is then given by
Math 401 - Differential Equation (Module 2) 1
∫ M1 (x) dx + ∫N1(y) dy = C where C is an arbitrary constant.
Example 1. Solve
(1 + x2)dy - xy dx = 0
Solution. Dividing by y(1 + x2) and transposing we get
dy xdx
=
y 1+ x2
Integrating both sides, we get
ln y = ½ ln (1 + x2) + ln C or ln y = ln C(1 + x2)½
Taking exponentials
y = C(1 + x2)½
The arbitrary constant was added in the form “ln C” to facilitate the final
representation.
B. Homogeneous differential equations. A differential equation of the form
M(x, y) dx + N(x, y) dy = 0
is said to be homogeneous if M and N are homogeneous functions of the same
degree in x and y. In this case we can write the equation in the form
dy M y
dx
=- N
= f( ).
x
This follows from the fact that M/N is a homogeneous function of degree zero
in x and y.
Such an equation can be transformed into an equation in which the
variables are separated by the substitution y = vx (or x = vy), where v is a
new variable.
Note. Differentiating y = vx gives dy = v dx + x dv, a quantity that must be
substituted for dy when vx is substituted for y.
Example 2. Solve
(x2 - y2)dx + 2xy dy = 0
Solution. We cannot separate the variables, but M(x, y) and N(x, y) are
homogeneous functions of degree 2. Substituting
y = vx and dy = v dx + x dv
we get
(1 - v2)dx + 2v(v dx + x dv) = 0
Separating the variables gives
2 vdv dx
=-
v 2+1 x
Math 401 - Differential Equation (Module 2) 2
Integrating we get
ln(v2 + 1) = - ln x + ln C
Taking exponentials we obtain
x(v2 + 1) = C
Finally, since v = y/x, this becomes
x2 + y2 = Cx
Reminders: Reason why the substitution y = vx transforms the equation
into one in which the variables are separable. The reason the substitution
y = vx transforms the equation into one in which the variables are separable
can be seen when the given equation is written in the form
dy M ( x , y)
=-
dx N (x , y)
If M(x, y) and N(x, y) are homogeneous functions of the same degree and one
substitutes vx for y one finds that the x’s all cancel out on the right side of
equation and the right side becomes a function in v alone i.e. the equation
takes the form
dy/dx = g(v)
Substituting dy = v dx + x dv then gives
v dx + x dv = g(v) dx
where the variables can be separated as
dv dx
=
g ( v )−v x
IV. Application of Learning: Find the general solution of the following differential
equations.
1. (y + 2)dx + (x- 2)dy = 0 5. (x + y )dx – xdy = 0
2. tan y dx + (x + 1)dy = 0 6. (y2 – xy) dx + x2dy = 0
3. dx + 3x2y2dy = 0 7. (x + y)dx + (x – y)dy = 0
4. y(1 – x)dx + x2dy = 0 8. dx – dy = y2dx + x2dy.
Math 401 - Differential Equation (Module 2) 3
Note: Write your answer in a yellow paper. In submitting your outputs, either you take a
picture and send it privately to my messenger account or you will pass a hardcopy and pass
it on the guardhouse. This will serve as your quiz.
Math 401 - Differential Equation (Module 2) 4
