CHAPTER 1
1.1
Definition and Classification of ODE
The ODE 𝑑𝑥̅
= 𝑓(𝑡, 𝑥̅) 𝑑𝑡 is called linear if 𝑓 has the form 𝑓(𝑡, 𝑥̅) = 𝐴(𝑡)𝑥̅ + 𝑏̅̅(𝑡) where 𝐴(𝑡) ∈ ℜ 𝑛×𝑛
and 𝑏̅̅ (𝑡) ∈ ℜ 𝑛
for all 𝑡 . A linear ODE is called homogeneous if 𝑏̅(𝑡) ≡ 0 and if 𝐴(𝑡) = 𝐴 (consant matrix), then we call it a linear ODE with constant coefficients. Linear ODE will be discussed in detail. The ODE – Eq. (1.2.1) is called autonomous if 𝑓 does not depend on 𝑡 . that is ODE has the form.
CHAPTER 2
1.2 Formation Of Differential Equation
Differential equations may be formed in practice from a consideration of the physical problems to which they refer. Mathematically, they can be occur when arbitrary constants are eliminated from a given function. Here are a few examples.
Example 1
Consider 𝑦 = 𝐴𝑠𝑖𝑛𝑥 + 𝐵𝑐𝑜𝑠𝑥 , where 𝐴 and 𝐵 are two arbitrary constants. If we differentiate, we get: 𝑑𝑦
= 𝐴cos𝑥 − 𝐵sin𝑥 𝑑𝑥 𝑑 2 𝑦
= −𝐴sin𝑥 − 𝐵scos𝑥 𝑑𝑥 2
Which is identical to the original equation, but with the sign changed.
i.e. 𝑑 2 𝑦 𝑑𝑥 2
= −𝑦
∴ 𝑑 2 𝑦 𝑑𝑥 2
+ 𝑦 = 0
This is a differential equation of the second order.
Example 2
Form a differential equation from the function 𝑦 = 𝑥 +
𝐴 𝑥
We have

∴ 𝑦 = 𝑥 +
𝐴 𝑥
= 𝑥 + 𝐴𝑥 −1
From the given equation,
𝐴 𝑥
∴ 𝑑𝑦 𝑑𝑥
= 1 − x(y−x) x 2 𝑑𝑦 𝑑𝑥
= 𝑦 − 𝑥
= 1 − 𝐴𝑥 −2 = 1 −
∴ 𝐴 = 𝑥(𝑦 − 𝑥) 𝑦 − 𝑥
= 1 − 𝑥
∴ 𝑥 𝑑𝑦
= 2x − 𝑦 𝑑𝑥
This is an equation of the first order.
= 𝑥 − 𝑦 + 𝑥) 𝑥
=
𝐴 𝑥 2
2𝑥 − 𝑦) 𝑥
Example 3
Form the differential equation for 𝑦 = 𝐴𝑥 2
We have 𝑦 = 𝐴𝑥 2 + 𝐵𝑥
+ 𝐵𝑥.
(1)
∴ 𝑑𝑦
= 2𝐴𝑥 + 𝐵 𝑑𝑥
∴ 𝑑 2 𝑦 𝑑𝑥 2
= 2𝐴
Substituting
∴ 𝐵 = 𝑑𝑦
− 𝑥 for 𝑑𝑦 𝑑𝑥
= 𝑥
2 𝐴 𝑑 2 𝑦
+ 𝐵 𝑑𝑥 2 𝑑𝑥 𝑑
2 𝑦 𝑑𝑥 2
Substituting for 𝐴 and 𝐵 in (1), we have 𝑦 = 𝑥 2 1
2 𝑥 2 2
= .
𝑑 𝑦
+ 𝑥.
d𝑦
2 𝑑𝑥 2 𝑑𝑥
∴ 𝑦 = 𝑥 𝑑𝑦 𝑑𝑥
− 𝑥
2
2
.
𝑑
2 𝑦 𝑑𝑥 2 and this is an equation of the second order. 𝑑
2 𝑑𝑥 𝑦
2
− 𝑥 2
+ 𝑥 ( d𝑦 𝑑𝑥
2
.
𝑑 𝑑𝑥 𝑦
2
− 𝑥 we 𝑑
2 𝑦 𝑑𝑥 2
) have
1.2
First order differential equation
1.2.1
Orders and Degrees
The definition of an ordinary differential equation implies that the most general first order equation may be written in the implicit form
𝐹(𝑥, 𝑦, 𝑦′) = 0 or if it can be solved for 𝑦′ , as the explicit equation.
(1) 𝑦′ = 𝑓(𝑥, 𝑦) . (2)
A solution of (1) or (2) is a differentiable function 𝑦 = 𝑔(𝑥) with the property that these equations become identities in x when y and y′ are replaced by 𝑔(𝑥) and 𝑔′(𝑥) , respectively. The graph of 𝑦 = 𝑔(𝑥) corresponding to a specific solution of (1) or (2) is called a solution curve, or an integral curve of the differential eqation.

It is sometimes convenient to classify a first order ordinary differential equation according to its degree. When (1) can be re-expressed as a polynomial equation in 𝑦′ (not necessarily in 𝑥 and 𝑦 as well) its degree is defined to be the degree m of that polynomial. Thus, the differential equation
(𝑦′)
3
3
= 1 + 𝑥 2 + sin2𝑦
Is of degree 3, because after clearing the radical the equation becomes
(𝑦′) 3 − (1 + 𝑥 2 + sin2𝑦) 2 = 0 ,
Which is a cubic in ′ .
It follows from the fundamental theorem of algebra (Theorem 1.3) that a polynomial of degree m in y′ can be solved to find m values of 𝑦′ , each of which depends on 𝑥 and 𝑦 , though not all of which are necessarily different. Thus, since solutions are obtained by integrating 𝑦′ , it then follows that an equation of the form (1), which can be re-expressed as a polynomial in y′ of degree m, can have at most m different solutions passing through any point at which its solutions exist. Clearly, such an equation can only have a unique solution when it is of degree 1.
By analogy, any first order ordinary differential equation (1) which is no expressible as a polynomial in 𝑦′ will be said to be of degree m if it has at most m different values of 𝑦′ at points where its solutions exist.
CHAPTER 2
2.1 Method1: By direct integration
If the equation can be arranged in the form 𝑑𝑦 𝑑𝑥 solved by direct integration.
Example 1
= 𝑓(𝑥) then the equation can be 𝑑𝑦
= 3𝑥 2 − 6𝑥 + 5 𝑑𝑥
Then 𝑦 = ∫(3𝑥 2 − 6𝑥 + 5)𝑑𝑥 = 𝑥 3 − 3𝑥 2 + 5𝑥 + 𝐶 i.e. 𝑦 = 𝑥 3 − 3𝑥 2 + 5𝑥 + 𝐶
As always, of course, the constant of integration must be included. Here it provides the one arbitrary constant which we always get when solving a first order differential quation.
Example 2
Solve
In 𝑥 𝑑𝑦
= 5𝑥 3 𝑑𝑥 this 𝑑𝑦
= 5𝑥 2
+ 4
+ 𝑑𝑥
4 𝑥 case,

So, 𝑦 =
5𝑥 3
3
+ 4ln𝑥 + 𝐶
Example 3
Find the particular solution of the equation 𝑒 𝑥 𝑑𝑦 𝑑𝑥 𝑥 = 0.
First rewrite the equation in the form 𝑑𝑦 𝑑𝑥
Then 𝑦 = ∫ 4𝑒 −𝑥 𝑑𝑥 == 4𝑒 −𝑥 + 𝐶
=
4 𝑒 𝑥
= 4𝑒
= 4, given that 𝑦 = 3 when
−𝑥 .
Knowing that when x=0, y=3, we can evaluate C in this case, so that the required particular solution is 𝑦 = −4𝑒 −𝑥 + 7.
2.2 Separable equations
First order differential equations capable of being written in either of the forms 𝑦′ = 𝑓(𝑥)𝑔(𝑥)
Or
𝐹(𝑥)𝐺(𝑦)𝑦′ + 𝑓(𝑥)𝑔(𝑥) = 0 are said to have separable variables, or simply to be separable equations. This is because the variables 𝑥 and 𝑦 occur in such a way that these differential equations can be rewritten with only a function of 𝑥 appearing on one side of the equations and a function of 𝑦 on the other. Thus (1) and (2) may be re-expressed as
1 𝑔(𝑦) 𝑦 ′ = 𝑓(𝑥), and
𝐺(𝑦) 𝑦 ′ 𝑔(𝑦)
= 𝑓(𝑥)
𝐹(𝑥)
,
In which the variables have been separated.
The significance of this separation of variables follows at once, because integrating both sides of these equations with respect to 𝑥, and remembering that 𝑦′ = 𝑑𝑦/𝑑𝑥, gives
1 𝑦 = ∫ 𝑑𝑦 𝑑𝑥 = ∫ 𝑓(𝑥)𝑑𝑥 𝑔(𝑦) 𝑑𝑥 and
𝐺(𝑦) 𝑑𝑦
∫ 𝑔(𝑦) 𝑑𝑥 𝑑𝑥 = − ∫ 𝑓(𝑥)
𝐹(𝑥) 𝑑𝑥
The rule for integration by substitution then permits the left-had sides of each of these results to be rewritten as an integral with respect to 𝑦, leading to the results

𝑦 = ∫
1 𝑔(𝑦) 𝑑𝑦 = ∫ 𝑓(𝑥)𝑑𝑥 + 𝑐 (7) and
∫
𝐺(𝑦) 𝑔(𝑦) 𝑓(𝑥) 𝑑𝑦 = − ∫
𝐹(𝑥) 𝑑𝑥 + 𝑐, (8)
It is essential the arbitrary constant c in (7) and (8) is introduced immediately the indicated integrations have been performed, and before the expressions are simplified to determine y explicitly. The constant c arises as a result of the integration, and only by including it at this stage can it enter into the solution in the correct manner.
2.2 Method2: By separating the variables
Example 1
Solve 𝑑𝑦
= 𝑑𝑥
2x y + 1
We can rewrite this as (𝑦 + 1) 𝑑𝑦
= 2𝑥 𝑑𝑥
Now integrate both sides with respect to 𝑥: 𝑑𝑦
(𝑦 + 1) ∫ 𝑑𝑥 = ∫ 2𝑥𝑑𝑥 𝑑𝑥 i.e.
(𝑦 + 1)𝑑𝑦 = ∫ 2𝑥𝑑𝑥 and this 𝑦 2
+ 𝑦 = 𝑥 2 + 𝐶
2
Example 2
Solve 𝑑𝑦
= (1 + 𝑥)(1 + 𝑦) 𝑑𝑥
1 𝑑𝑦
= 1 + 𝑥
1 + 𝑦 𝑑𝑥
Integrating both sides with respect to 𝑥: gives

1
∫
1 + 𝑦 𝑑𝑦 𝑑𝑥 = ∫(1 + 𝑥) 𝑑𝑥 𝑑𝑥
∴
1
∫
1 + 𝑦 𝑑𝑦 = ∫(1 + 𝑥) 𝑑𝑥 𝑥 2 ln(1 + 𝑦) = 𝑥 + + 𝐶
2 the method depends on our being able to express the given equation in the form F(y). 𝑑𝑦 𝑑𝑥
= 𝑓(𝑥). If this can be done, the rest is then easy, for we have 𝑑𝑦
∫ 𝐹(𝑦).
𝑑𝑥 = ∫ 𝑓(𝑥) 𝑑𝑥 𝑑𝑥
And we than continue as in the examples.
Example 3
Solve 𝑑𝑦
=
1+𝑦 𝑑𝑥 (1) 𝑑𝑥 2+𝑥
This can be written as
1
1 + 𝑦 𝑑𝑦 𝑑𝑥
=
1
2 + 𝑥
Integrate both sides with respect to 𝑥:
1
∫ 𝑑𝑦 𝑑𝑥 = ∫
1 + 𝑦 𝑑𝑥
∴
1
2 + 𝑥
1
∫
1 + 𝑦 𝑑𝑦 𝑑𝑦 = ∫ 𝑑𝑥 𝑑𝑥
1
2 + 𝑥 𝑑𝑥
∴ ln(1 + 𝑦) = ln(2 + 𝑥) + 𝐶
It is convenient to write the constant 𝐶 as the logarithm of some other constant 𝐴: ln(1 + 𝑦) = ln(2 + 𝑥) + ln𝐴 = ln𝐴(2 + 𝑥)
∴ 1 + 𝑦 = 𝐴(2 + 𝑥)
Note that we can, in practice, get the given equation (1) to the form of the equation in (2) by a simple routine, thus: 𝑑𝑦
=
1 + 𝑦 𝑑𝑥 2 + 𝑥
First multiply across by 𝑑𝑥:
1 + 𝑦 𝑑𝑦 = 𝑑𝑥
2 + 𝑥
Now collect the ‘𝑦-factor’ with the d𝑦 on the left, i.e. divide by (1 + 𝑦):
1 1 𝑑𝑦 = 𝑑𝑥
1 + 𝑦 2 + 𝑥

Finally, add the integral signs:
1
∫
1 + 𝑦
1 𝑑𝑦 = ∫
2 + 𝑥 𝑑𝑥
And then continue as before.
This is purely a routine which enables us to sort out the equation algebraically, the whole of the work being done in one line. Notice, however, that the RHS of the given equation must be expressed as ‘𝑥-factors’ and ‘yfactors’.
Example 4
Solve 𝑑𝑦
= 𝑦 2 𝑥 2
+ 𝑥𝑦 2 𝑦 − 𝑥 2 𝑑𝑥
First express the RHS in ‘x-factors’ and dy on the LHS and the ‘x-factors’ and dx on the RHS. 𝑦 − 1 𝑑𝑦 = 𝑦 2
We now add the integral signs: 𝑦 − 1
∫ 𝑦 2
1 + 𝑥 𝑑𝑦 = ∫ 𝑥 2 𝑥 2 𝑑𝑥
1 + 𝑥 𝑑𝑥
And complete the solution:
∫ { 𝑦 − 1 𝑦 2
} 𝑑𝑦 = ∫ {
1 + 𝑥
} 𝑑𝑥 𝑥 2
1
∴ ln𝑦 + = ln𝑥 − 𝑦
Here is another.
1 𝑥
+ 𝐶
Example 5
Solve 𝑑𝑦 𝑑𝑥
= 𝑦 2 − 1 𝑥
Rearranging, we have 𝑦 2 𝑑𝑦 =
− 1 𝑑𝑥 𝑥

1 𝑦 2 − 1
1
∫ 𝑦 2 − 1 𝑑𝑦 =
1 𝑥 𝑑𝑥
1 𝑑𝑦 = ∫ 𝑑𝑥 𝑥
Which gives
∫ 𝑦 2
1
− 1 𝑑𝑦 = ∫
1 𝑥 𝑑𝑥
1
2 ln 𝑦 − 1 𝑦 + 1
= ln𝑥 + 𝐶
∴ ln 𝑦 − 1 𝑦 + 1
= ln𝑥 + ln𝐴
∴ 𝑦 − 1
= 𝐴𝑥 2
∫(𝑦 2 𝑦 + 1 y-1=A𝑥 2 (𝑦 + 1)
You see they are all done in the same way. Now hear is on for you to do:
Example 6
Solve 𝑥𝑦 𝑑𝑦 𝑑𝑥
= 𝑦 2 + 1 𝑦 + 1
First of all, rearrange the equation into the form:
𝐹(𝑦)𝑑𝑦 = 𝑓(𝑥)𝑑𝑥
I.e. arrange the ‘y-factors’ and dy on the LHS and the ‘x-factors’ and dx on the
RHS.
∴ 𝑦(𝑦 + 1)𝑑𝑦 = 𝑥 2 + 1 𝑑𝑥 𝑥
Because 𝑥𝑦𝑑𝑦 = 𝑥 2 + 1 𝑑𝑥 𝑦 + 1
∴ 𝑦(𝑦 + 1)𝑑𝑦 =
So we now have 𝑥
2
+1 𝑑𝑥 𝑥
+ 𝑦)d𝑦 = ∫ (1 +
1 𝑥
) d𝑥

𝑥 3 𝑥 2 𝑥 2
+ = + ln𝑥 + 𝐶
3 2 2 𝑑𝑦
Provided that the RHS of the equation = 𝑓(𝑥, 𝑦) can be separated into ‘x𝑑𝑥 factors’ and ‘y-factors’, the equation can be solved by the method of separating the variable.
Example 7
Solve 𝑥 𝑑𝑦 𝑑𝑥
= 𝑦 + 𝑥𝑦
Solution 𝑥 𝑑𝑦 𝑑𝑥
= 𝑦(1 + 𝑥) 𝑥𝑑𝑦 = 𝑦(1 + 𝑥)𝑑𝑥 ∴ 𝑑𝑦 𝑦
=
1 + 𝑥 𝑑𝑥 𝑥
∴
∫
1 𝑦 𝑑𝑦 = ∫ (
1 𝑥
+ 1) 𝑑𝑥
∴ ln𝑦 = ln𝑥 + 𝑥 + 𝐶
At this stage we have eliminated the derivatives and so we have solved the equation.
However, we can express the result in a neater form, thus: ln𝑦 − ln𝑥 = 𝑥 + 𝐶
∴ { 𝑦 𝑥
} = 𝑥 + 𝐶
∴ 𝑦
= 𝑒 𝑥+𝑐 = 𝑒 𝑥 . 𝑒 𝑐 𝑥
Let 𝐴 = 𝑒 𝑐
∴ 𝑦
= 𝐴𝑒 𝑥 𝑥
∴ 𝑦 = 𝐴𝑥𝑒 𝑥
Example 8
Solve 𝑦tan𝑥 𝑑𝑦
= (4 + 𝑦 2 )sec 2 𝑥 𝑑𝑥
First separate the variables, i.e. arrange the ‘𝑦-factors’ and d𝑦 on one side and the ‘𝑥-factors’ and d𝑥 on the other. 𝑦𝑑𝑦
4 + 𝑦 2
= 𝑠𝑒𝑐 2 𝑥𝑑𝑥 𝑡𝑎𝑛𝑥

Applying the integration symbol, we have:
∫ 𝑦𝑑𝑦
4 + 𝑦 2
= ∫ 𝑠𝑒𝑐 2 𝑥𝑑𝑥 𝑡𝑎𝑛𝑥
And we have,
1 ln(4 + 𝑦 2 ) = ln (tan𝑥)
2
Example 9
Solve the differential equation 𝑦′ = 𝑥 2 (1 + 𝑦 2 ).
Solution
Separating the variables and integrating both sides as in (7) gives
1
∫
1 + 𝑦 2 𝑑𝑦 = ∫ 𝑥 2 𝑑𝑥
Which after the integrations have been performed shows tan −1 𝑦 =
1
3 𝑥
3
+𝑐 where c is an arbitrary constant. Thus the explicit general solution is 𝑦 = tan (
1
3 𝑥
3
+𝑐)
Example 10
Solve the differential equation 𝑥 2 𝑦 2 𝑦 ′ − (1 + 𝑥)(1 + 𝑦) = 0.
Solution
Separating the variables and integrating both sides as in (8) gives 𝑦 2
∫
1 + 𝑦 𝑑𝑦 ∫
1 + 𝑥 𝑥 2 𝑑𝑥
Which after simplification becomes
∫ (𝑦 − 1 +
1
1 + 𝑦
1
) 𝑑𝑦 = ∫ ( 𝑥 2
+
1 𝑥
) 𝑑𝑥
Performing the indicated integrations gives
1 𝑦 2 − 𝑦 + ln|1 + 𝑦| = −
1
+ ln|𝑥| + 𝑐,
2 𝑥 with an arbitrary constant 𝑐. This is an implicit equation for 𝑦 and cannot be further simplified.
When first order differential equation is required to satisfy an initial condition 𝑦 = 𝑦
0
when 𝑥 = 𝑥
0
, written

𝑦(𝑥
0
) = 𝑦
0 the combination of the equation and the initial condition is called an initial value problem. This name arises because in many physical situations the independent variable is the time, and a condition of this type specifies how a solution is to start. The following examples of physical problems leading to separable equations illustrate how initial conditions arise. They also show how the initial condition must be used to determine the appropriate value of the arbitrary constant in the general solution if it is to satisfy the initial value problem.
Example 11
When a body cools in air subject to Newton’s law of cooling, its surface temperature 𝑇(𝑡) at t satisfies the differential equation 𝑑𝑇 𝑑𝑡
= −𝜆(𝑇 − 𝑇
0
), (10) where 𝜆 is a constant depending on the body and 𝑇
0
is the constant ambient air temperature. We shall now solve an initial value problem for this equation in which the initial conditions is 𝑇(0) = 𝑇
1 in (10) and integrating both sides gives
. Separating the variables
1
∫
𝑇 − 𝑇
0 𝑑𝑇 = −𝜆𝑑𝑡
Or ln(𝑇 − 𝑇
0 where c is an arbitrary constant.
) = −𝜆𝑡 + 𝑐,
Taking the exponential of both sides of this equation then shows
𝑇 − 𝑇
0
= 𝐴𝑒 −𝜆𝑡
Where 𝐴 = 𝑒 𝑐
,
. As 𝑐 was an arbitrary constant, it follows that 𝐴 must be an arbitrary positive constant. No useful purpose is served by performing arithmetic on arbitrary constants, so from this point onwards we shall work with 𝐴 rather than 𝑐. The general solution of (10) is thus
𝑇 = 𝐴𝑒 −𝜆𝑡 + 𝑇
0
To solve the initial value problem it is necessary to choose 𝐴 so that 𝑇(0) =
𝑇
1
. Setting 𝑇 = 𝑇
1
and 𝑡 = 0 in the general solution shows
𝑇
1
= 𝐴 + 𝑇
0
or 𝐴 = 𝑇
1
− 𝑇
0

The required solution of the initial value problem is thus
− 𝑇
0
)𝑒 −𝜆𝑡
For 𝑡 ≥ 0.
𝑇 = (𝑇
1
+ 𝑇
0
This solution shows that with passage of time the body surface temperature decays exponentially to the ambient air temperature 𝑇
0
.
Example 12. The logistic equation
The study of population growth, the rate of learning, the spread of information, the acceptance of newly marketed products and various other analogous situations is often based on the mathematical model called the logistic equation, or the equation of inhibited growth. 𝑑𝑃 𝑑𝑡
= 𝑎𝑃 (1 −
𝑃
𝐿
) (12) where 𝑎 and 𝐿 are positive constants.
This equation is derived on the assumption that a population 𝑃 cannot grow beyond some known posiive valueL, and that the rate of growth of 𝑃 is
𝑃 proportional both to 𝑃 and to (1 − ) the fraction of growth which remains
𝐿 possible. The positive number 𝑎 in (12) is proportionality constant. Let us solve an initial value problem for this equation in which the initial condition
𝑃(0) = 𝑃
0
with 0 < 𝑃
0
< 𝐿.
Separating the variables and integrating both sides of (12) give
∫
1
𝑃 (1 −
𝑃
𝐿) 𝑑𝑃 = ∫ 𝑎dt which after simplification by means of partial fractions becomes
∫ (
1
𝑃
+
1
𝐿 − 𝑃
) 𝑑𝑃 = ∫ 𝑎dt (13)
Carrying out the indicated integration we find
𝑃 ln (
𝐿 − 𝑃
) = 𝑎𝑡 + 𝑐, where c is an arbitrary constant. Taking the exponential of both sides of this equation shows
𝑃
= 𝐴𝑒 𝑎𝑡
𝐿 − 𝑃
Here again we have set 𝐴 = 𝑒 𝑐 , with A an arbitrary positive constant.
Rearranging terms brings the general solution of the logistic equation to the form
𝑃 =
𝐿
1 + 𝐴𝑒 −𝑎𝑡
(14) for 𝑡 ≥ 0.

To solve the stated initial value problem it is now necessary to choose A so that 𝑃(0) = 𝑃
0
. Setting 𝑃 = 𝑃
0
and 𝑡 = 0 in (14) gives
𝐴 =
𝑃
0
𝐿 − 𝑃
0
Combining this results with (14) finally gives for the solution of the initial value problem
𝑃 =
1 + (
𝐿
𝑃
0
𝐿
− 1) 𝑒 −𝑎𝑡
for 𝑡 ≥ 0
A simple calculation establishes that the curve has a point of inflection at 𝑄, where 𝑃 = 1
2
𝐿 and 𝑡 =
1 𝑎
𝐿 ln (
𝑃
0
− 1).
2.3 Method 3: Homogeneous equations – by substituting 𝒚 = 𝒗𝒙
Here is an equation: 𝑑𝑦 𝑑𝑥
= 𝑥 + 3𝑦
2𝑥
This looks simple enough, but we find that we cannot express the RHS in the form of
‘𝑥-factors’ and ‘𝑦-factors’, so we cannot solve by the method of separating the variables. In this case, we make the substitution 𝑦 = 𝑣𝑥, where 𝑣 is a function of 𝑥.
So 𝑦 = 𝑣𝑥. Differentiate with respect to 𝑥 (using the product rule): 𝑑𝑦
= 𝑣. 1 + 𝑥 𝑑𝑣
= 𝑣 + 𝑥 𝑑𝑣 𝑑𝑥 𝑑𝑥 𝑑𝑥
Also, 𝑥 + 3𝑦 𝑥 + 3𝑣𝑥 1 + 3𝑣
= =
2𝑥 2𝑥 2
The equation now becomes 𝑣 + 𝑥 𝑑𝑣 𝑑𝑥
=
1 + 3𝑣
2
∴
= 𝑑𝑣 1 + 3𝑣 𝑥 = 𝑑𝑥 2
1 + 3𝑣 − 2𝑣
=
2
− 𝑣
∴
1 + 𝑣
2 𝑥 𝑑𝑣 𝑑𝑥
=
1 + 𝑣
2

The given equation is now expressed in terms of 𝑣 and 𝑥 and in this form we find that we can solve by separating the variables. Here goes:
2
∫
1 + 𝑣
∴ 2ln(1 + 𝑣) = ln𝑥 + 𝐶 = ln𝑥 + ln𝐴 𝑑𝑣 = ∫
(1 + 𝑣) 2
1 𝑥
= 𝐴𝑥 𝑑𝑥
But 𝑦 = 𝑣𝑥 ∴ 𝑣 = { 𝑦 𝑥
} ∴ (1 +
Which gives (𝑥 + 𝑦) 2 = 𝐴𝑥 3 𝑥 𝑦
)
2
= 𝐴𝑥
Note that 𝑥 + 3𝑦
2𝑥 is an example of a homogeneous differential equation.
This is determined by the fact that the total degree in 𝑥 and 𝑦 for each of the terms involved is the same (in this case, of degree 1). The key to solving every homogeneous equation is to substitute 𝑦 = 𝑣𝑥 where 𝑣 is a function of 𝑥. This converts the equation into a form which we can solve by separating the variables.
Example 1
Solve 𝑑𝑦 𝑑𝑥
= 𝑥 2 + 𝑦 2 𝑥𝑦
Here, all terms of the RHS are of degree 2, i.e. the equation is homogeneous.
∴ We substitute 𝑦 = 𝑣𝑥 (where 𝑣 is a function of 𝑥).
∴ 𝑑𝑦 𝑑𝑥 and
= 𝑣 + 𝑥 𝑑𝑣 𝑑𝑥 𝑥 2 + 𝑦 2 𝑥 2 + 𝑣 2 𝑦 2 1 + 𝑣 2 𝑥𝑦
= 𝑣𝑥 2
= 𝑣
The equation now becomes: 𝑣 + 𝑥 𝑑𝑣 𝑑𝑥
=
1 + 𝑣 2 𝑣
∴ 𝑥
= 𝑑𝑣
= 𝑑𝑥
1 + 𝑣
1 + 𝑣
2 𝑣
− 𝑣
2
2 𝑣
− 𝑣
=
∴
1 𝑣 𝑥 𝑑𝑣 𝑑𝑥
=
1 𝑣

Separating the variables we have
1
∫ 𝑣𝑑𝑣 = ∫ 𝑑𝑥 𝑥
∴ 𝑣 2
= 𝑙𝑛𝑥 + 𝐶
2
All that remains now is to express 𝑣 back in terms of 𝑥 and 𝑦. the substitution we used was 𝑦 = 𝑣𝑥
∴ 𝑣 = 𝑦 𝑥
∴
1 𝑦
2
2
( 𝑦 𝑥
)
2
= 2𝑥
= ln𝑥 + 𝐶
2 (ln𝑥 + 𝐶)
Example 2
Solve 𝑑𝑦 𝑑𝑥
=
2𝑥𝑦 + 3𝑦 2 𝑥 2 + 2𝑥𝑦
Solution
This is a homogeneous equation, because the degree of each term is the same. We substitute 𝑦 = 𝑣𝑥, where 𝑣 is a function of 𝑥.
Thus, 𝑑𝑦 𝑑𝑥
=
2𝑥𝑦 + 3𝑦 2 𝑥 2 + 2𝑥𝑦
2𝑥𝑦 + 3𝑦 𝑥 2 + 2𝑥𝑦
2
= 𝑑𝑦
= 𝑣 + 𝑥 𝑑𝑥
2𝑣𝑥 2 + 3𝑣 𝑥 2 + 2𝑣𝑥 𝑑𝑣
2 𝑑𝑥 𝑥 2
2
=
2𝑣 + 3𝑣
1 + 2𝑣
So that 𝑑𝑦 2𝑣 + 3𝑣 2 𝑣 + 𝑥 = 𝑑𝑥 1 + 2𝑣
Now take the single 𝑣 over to the RHS and simplify, giving: 𝑥
= 𝑑𝑣 𝑑𝑥
=
2𝑣 + 3𝑣
1 + 2𝑣
2𝑣 + 3𝑣 2
2
− 𝑣
− 𝑣 − 2𝑣 2 𝑥 𝑑𝑣 𝑑𝑥
1 + 2𝑣 𝑣 + 𝑣
=
2
1 + 2𝑣
Separating the variables, we have:
2
∫
1 + 2𝑣 𝑣 + 𝑣 2 𝑑𝑣 = ∫
1 𝑥 𝑑𝑥
Integrating both sides, we can now obtain the solution in terms of 𝑣 and 𝑥.

ln (𝑣 + 𝑣 2 ) = lnx + C = lnx + lnA
∴ 𝑣 + 𝑣 2 = 𝐴𝑥
We express 𝑣 back in terms of 𝑥 and 𝑦.
Since 𝑦 = 𝑣𝑥, then 𝑣 = 𝑦 𝑥
and we have 𝑦
+ 𝑥
Therefore
𝑥𝑦 + 𝑦 2 = 𝐴𝑥 3 𝑦 𝑥
2
2
= 𝐴𝑥
Example 3
Solve
(𝑥 2 + 𝑦 2 ) 𝑑𝑦 𝑑𝑥
= 𝑥𝑦
Solution
(𝑥 2 + 𝑦 2 ) 𝑑𝑦 𝑑𝑥
= 𝑥𝑦 𝑑𝑦 𝑑𝑥
= 𝑥𝑦
(𝑥 2 + 𝑦 2 )
Let 𝑦 = 𝑣𝑥
Therefore, 𝑑𝑦 𝑑𝑥
= 𝑣 + 𝑥 𝑑𝑣 𝑑𝑥 and 𝑥𝑦 𝑣𝑥 2 𝑥 2 + 𝑦 2
= 𝑥 2 + 𝑣 2 𝑥 2
= 𝑣
1 + 𝑣 2
Therefore 𝑣 + 𝑥 𝑑𝑣 𝑑𝑥
+ 𝑣
1 + 𝑣 2 𝑥 𝑑𝑣 𝑑𝑥
= 𝑣
1 + 𝑣 2
− 𝑣 𝑥 𝑑𝑣 𝑑𝑥
= 𝑣 − 𝑣 − 𝑣 3
1 + 𝑣 2
=
−𝑣 3
1 + 𝑣 2
Integrating now we have
∫
1 + 𝑣 2 𝑣 3 𝑑𝑣 = − ∫ 𝑑𝑥 𝑥 therefore ln 𝐾 𝑣𝑥 =
1
2𝑣 2
But 𝑣 = 𝑦 𝑥

Therefore ln 𝐾𝑦 = 𝑥
2
2𝑦 2
Exercises
Solve the following 𝑑𝑦
1.
(𝑥 − 𝑦) 𝑑𝑥
= 𝑥 + 𝑦
2.
2𝑥 2 𝑑𝑦 𝑑𝑥
3.
(𝑥 2
= 𝑥
+ 𝑥𝑦)
2 𝑑𝑦 𝑑𝑥
+ 𝑦 2
= 𝑥𝑦 − 𝑦 2
2𝑦 2 ln𝐾𝑦 = 𝑥 2