• General and Standard Forms of linear first-order ordinary differential equations.
• Theory of solving these ODE’s.
• Direct Method of solving linear first-order ODE’s.
• Examples.

𝑎
1 𝑥 .
𝑑𝑦 𝑑𝑥
+ 𝑎
0 𝑥 . 𝑦 = 𝑔(𝑥)
• It is linear, so there are no functions of 𝑦 or any of its derivatives.
• The highest order is 𝑑𝑦 𝑑𝑥
.
• There is a function of 𝑥 represented by 𝑔(𝑥) , though this function may also be equal to 0.

• The general form of a linear first-order ODE is 𝒂
𝟏 𝒙 .
𝒅𝒚 𝒅𝒙
+ 𝒂
𝟎 𝒙 . 𝒚 = 𝒈(𝒙)
• In this equation, if 𝑎
1 𝑥 = 0 , it is no longer an differential equation and so 𝑎 if 𝑎
0
1 𝑥
cannot be 0; and 𝑥 = 0 , it is a variable separated ODE and can easily be solved by integration, thus in this chapter 𝑎
0 𝑥 cannot be 0.
If 𝑔 𝑥 = 0 , it is called a Homogenous Equation, and can easily be solved by separating the variables, thus in this chapter g(𝑥) is generally not 0. If 𝑔(𝑥) ≠ 0 it is a non-homogenous equation.

By dividing both sides of this general form by 𝑎
1
(𝑥) we get the standard form, which is much more useful for solving it: 𝒅𝒚 𝒅𝒙
+ 𝑷 𝒙 𝒚 = 𝒇(𝒙) where 𝑃 𝑥 = 𝑎
0 and f 𝑥 = 𝑔 𝑥 / 𝑎
1 𝑥 / 𝑎
1 𝑥 𝑥
There is a very important theory behind the solution of differential equations which is covered in the next few slides. For a review of the direct method to solve linear first-order differential equations, jump ahead to the direct method on slide
14.

• Solving differential equations is based on the property that the solution 𝑦(𝑥) can be represented as 𝑦 𝑐 𝑥 + 𝑦 𝑦 𝑐 𝑥 is the solution of the homogenous equation 𝑑𝑥
𝑃 𝑥 𝑦 = 0 and 𝑦 𝑝
(𝑥) is a particular solution of the entire non-homogenous equation 𝑑𝑦 𝑑𝑥
+ 𝑃 𝑥 𝑦 = 𝑓 𝑥 . 𝑝
(𝑥) 𝑑𝑦
+
, where

• There are 2 important points to note:
1.
Clearly, if the original problem is a homogenous equation, 𝑦 𝑝 𝑥 = 0 and the solution is 𝑦 = 𝑦 𝑐 𝑥 .
2.
We can see that 𝑦 𝑐 𝑥 = 0 is always valid. However this is known as the trivial solution and we are only interested in finding out all non-trivial solutions.

𝑦 𝑐
(𝑥)
The equation 𝑑𝑦
+ 𝑃 𝑥 𝑦 = 0 can be re-written as 𝑑𝑦 𝑑𝑥 𝑦
− 𝑃 𝑥 𝑑𝑥 with simple algebra. Integrating both sides,
= ln 𝑦 = − 𝑃 𝑥 𝑑𝑥
∴ 𝑦 = 𝑒 − 𝑃 𝑥 𝑑𝑥+𝑐
Where c is the constant of integration.

𝑦 𝑐 𝑥
We can rewrite this expression for 𝑦 𝑐
= 𝑐𝑒 𝑦 𝑐
− 𝑃 𝑥 𝑑𝑥 𝑥 as
= 𝑐𝑦
1 where 𝑐 is the parameter which can have any real value, and all the obtained functions will be solutions of the associated homogenous equation of the original differential equation; and 𝑦
1
= 𝑒 − 𝑃 𝑥 𝑑𝑥

𝑦 𝑐 𝑥
Let us consider the homogenous equation
3𝑦 ′ + sin 𝑥 + 3𝑒 𝑎𝑥 𝑦 = 0 where a is a constant. The equation can be re-written as: 𝑦′ 𝑦
= −
1
3 sin 𝑥 + 3𝑒 𝑎𝑥
Integrating both sides with respect to x , we get ln 𝑦 = −
1
3
sin 𝑥 𝑑𝑥 + 3𝑒 𝑎𝑥 𝑑𝑥 = cos 𝑥
3
− 𝑒 𝑎𝑥 𝑎
+ 𝑐
1 𝑐 previous slide.
∴ 𝑦 = 𝑐𝑒 cos 𝑥
3
− 𝑒 𝑎𝑥 𝑎 = 𝑦 𝑐
Where 𝑐 is any arbitrarily chosen parameter. Alternatively, this answer for 𝑦 𝑥 can be obtained from the formula on the

𝑦 𝑝
(𝑥)
To find 𝑦 𝑝 𝑦 𝑝
(𝑥) we use the method of Variation of Parameters and make the assumption that it is of the form 𝑢(𝑥)𝑦
1 unknown function and 𝑦
1
− 𝑃 𝑥 𝑑𝑥
(𝑥) where 𝑢(𝑥) is an
. Substituting this form of
(𝑥) into the Standard form of the equation, we get 𝑢𝑦 ′
1
= 𝑒
+ 𝑦
1 𝑢 ′ + 𝑃 𝑥 𝑢𝑦
1
= 𝑓(𝑥)
Rearranging and substituting and therefore 𝑑𝑦
1 𝑑𝑥
+ 𝑃 𝑥 𝑦
1
= 0 , we get 𝑢 = 𝑓(𝑥) 𝑦
1
(𝑥) 𝑑𝑥 𝑦 𝑝 𝑥 = 𝑢 𝑥 𝑦
1 𝑥 = 𝑒 − 𝑃 𝑥 𝑑𝑥 𝑒 𝑃 𝑥 𝑑𝑥 𝑓(𝑥)𝑑𝑥

𝑦(𝑥)
Going back to the original equation 𝑦 𝑥 = 𝑦 𝑐 substitute and get 𝑥 + 𝑦 𝑝
(𝑥) we 𝑦 𝑥 = 𝑒 − 𝑃 𝑥 𝑑𝑥 (𝑐 + 𝑒 𝑃 𝑥 𝑑𝑥 𝑓 𝑥 𝑑𝑥)
Which is the entire solution for the differential equation that we started with.
Using this equation we can now derive an easier method to solve linear first-order differential equation.

The previous equation can be re-written as: 𝑦 𝑥 𝑒 𝑃 𝑥 𝑑𝑥 = 𝑐 + 𝑒 𝑃 𝑥 𝑑𝑥 𝑓 𝑥 𝑑𝑥
Differentiating both sides, we get 𝑦 ′ 𝑒 𝑃 𝑥 𝑑𝑥 + 𝑃 𝑥 𝑦𝑒 𝑃 𝑥 𝑑𝑥 = 𝑒 𝑃 𝑥 𝑑𝑥 𝑓 𝑥
Which is clearly the Standard Form multiplied with 𝑒 𝑃 𝑥 𝑑𝑥
. Thus we can clearly see following these steps in reverse, we get an easy way to solve these differential equations.

The trick is to represent the left side of the Standard Form of the equation in the form of the product rule, while the right side is some function of the independent variable ( 𝑥 ). To do so, we multiply the entire differential equation with the Integrating
Factor 𝑒 𝑃 𝑥 𝑑𝑥
to get the equation: 𝑑𝑦 𝑑𝑥 𝑒 𝑃 𝑥 𝑑𝑥 + 𝑃 𝑥 𝑦𝑒 𝑃 𝑥 𝑑𝑥 = 𝑒 𝑃 𝑥 𝑑𝑥 𝑓 𝑥

We can see that the left side is in the form of the product rule: 𝑑 𝑑𝑥 𝑦 𝑥 . 𝑒 𝑃 𝑥 𝑑𝑥 = 𝑒 𝑃 𝑥 𝑑𝑥 𝑓 𝑥
Integrating this equation with respect to 𝑥 , we get 𝑦 𝑥 . 𝑒 𝑃 𝑥 𝑑𝑥 = 𝑒 𝑃 𝑥 𝑑𝑥 𝑓 𝑥 𝑑𝑥 + 𝑐
Which can then easily be solved for 𝑦(𝑥) .

1. Solve the differential equation 𝑑𝑦 𝑑𝑥
+ 𝑦 = 𝑒 3𝑥
.
SOLUTION: Since it is already in the standard form, we can directly see that 𝑃 𝑥 = 1 .
Thus, the Integrating factor is: 𝑒 𝑃 𝑥 𝑑𝑥 = 𝑒 1𝑑𝑥 = 𝑒 𝑥
Multiplying the equation with this, we get 𝑑𝑦 𝑑𝑥 𝑒 𝑥 + 𝑦𝑒 𝑥 = 𝑒 3𝑥 . 𝑒 𝑥

∴ 𝑑 𝑑𝑥 𝑦. 𝑒 𝑥 = 𝑒 4𝑥
Integrating this equation with respect to 𝑥 , y. 𝑒 𝑥 = 𝑒 4𝑥 𝑑𝑥 = 𝑒 4𝑥
4
+ 𝑐
∴ 𝑦 = 𝑒 3𝑥
4
+ 𝑐𝑒 −𝑥
Which is the desired solution for the differential equation. Here, 𝑐 is any arbitrary real number.

2. Solve: cos 2 𝑥 sin 𝑥 𝑑𝑦 𝑑𝑥
+ cos 3 𝑥 𝑦 = 1
SOLUTION: We first change it from the General Form to the Standard
Form. 𝑑𝑦 𝑑𝑥
+ cos (𝑥) sin (𝑥) 𝑦 = cos 2
1 𝑥 sin 𝑥
The integrating factor is 𝑒 cos (𝑥) sin (𝑥)𝑑𝑥 = 𝑒 ln |sin (𝑥)| = sin (𝑥)
Multiplying the standard form with sin (𝑥) , we get: sin (𝑥) 𝑑𝑦 𝑑𝑥
+ cos (𝑥)𝑦 = cos
1
2 𝑥

∴ 𝑑 𝑑𝑥 𝑦. sin 𝑥 = cos
1
2 (𝑥)
∴ 𝑦. sin 𝑥 = sec 2 𝑥 𝑑𝑥 = tan 𝑥 + 𝑐
∴ 𝑦 𝑥 =
1 cos (𝑥)
+ 𝑐 sin (𝑥)
Alternatively we can write:
∴ 𝑦 𝑥 = sec 𝑥 + 𝑐. csc (𝑥) which is the desired solution.

3. Solve: 𝑦 𝑑𝑥 𝑑𝑦
− 𝑥 = 2𝑦 2 ; 𝑦 1 = 5
SOLUTION: This sum covers 2 important new ideas. Firstly, as can be seen from the given
ODE, 𝑥 is the dependent variable and not 𝑦 .
Secondly, an Initial Value of 𝑦 = 5 when 𝑥 = 1 is given to find a specific value of the arbitrary parameter 𝑐 .
Keeping in mind that x is dependent and so we will integrate with respect to y, we convert to Standard Form as: 𝑑𝑥 𝑑𝑦
−
1 𝑦 𝑥 = 2𝑦
The integrating factor is 𝑒
−
1 𝑦𝑑𝑦 = 𝑒 −ln |𝑦| =
1 𝑦
Multiplying the standard form with
1 𝑦
, we get:
1 𝑑𝑥 1 𝑦
∴ 𝑑𝑦 𝑑 𝑑𝑦
−
1 𝑦 𝑦 2 𝑥 = 2
. 𝑥 = 2
Integrating with respect to 𝑦 ;
1 𝑦
. 𝑥 = 2𝑦 + 𝑐
∴ 𝑥 = 2𝑦 2 + 𝑐𝑦

This equation is the solution of the ODE that we started off with.
However we are not done yet. We will now use the given Initial Value to solve for a particular value of 𝑐 for this problem.
Substituting 𝑦 = 5 and 𝑥 = 1 into the solution 𝑥 = 2𝑦
1 = 2. 5 2
∴ 𝑐 = −
+ 𝑐. 5
∴ 1 − 50 = 5𝑐
49
5
2 + 𝑐𝑦 , we get:
Substituting this value for 𝑐 back into the solution equation, we get our complete final solution: 𝑥 = 2𝑦 2 −
49
5 𝑦

4. Solve 𝑑𝑇 𝑑𝑡 constants.
= 𝑘(𝑇 − 𝑇 𝑚
) ; 𝑇 0 = 𝑇
0
where 𝑘, 𝑇 𝑚
and 𝑇
0
are
SOLUTION: Rewriting in standard form: 𝑑𝑇 𝑑𝑡
− 𝑘𝑇 = −𝑘𝑇 𝑚
Integrating Factor: 𝑒 −𝑘𝑑𝑡 = 𝑒 −𝑘𝑡
.
Multiplying with the Integrating Factor, 𝑒 −𝑘𝑡
∴ 𝑑𝑇 𝑑𝑡 𝑑 𝑑𝑡
− 𝑘𝑒
𝑇. 𝑒
∴ 𝑇. 𝑒 −𝑘𝑡
−𝑘𝑡
−𝑘𝑡
𝑇 = −𝑘𝑒
= −𝑘𝑒
= 𝑇 𝑚
∴ 𝑇 = 𝑇 𝑚 𝑒 −𝑘𝑡
+ 𝑐𝑒
−𝑘𝑡 𝑘𝑡
−𝑘𝑡
𝑇
+ 𝑐 𝑚
𝑇 𝑚

Putting in the given initial conditions,
∴ 𝑇
0
= 𝑇
∴ 𝑐 = (𝑇 𝑚
0
+ 𝑐𝑒
− 𝑇 𝑚
0
)
Substituting back into the original solution equation,
𝑇 = 𝑇 𝑚
+ (𝑇
0
− 𝑇 𝑚
)𝑒 𝑘𝑡 which is our answer.

5.
Solve 𝑦 ′ + (tan 𝑥)𝑦 = cos 2 𝑥
SOLUTION: Integrating Factor is 𝑒 (tan 𝑥)𝑑𝑥 = sec 𝑥
Multiplying, we get
(sec 𝑥)𝑦
∴
′ 𝑑 𝑑𝑥
+ (sec 𝑥 tan 𝑥)𝑦 = cos 𝑥
[𝑦. sec 𝑥] = cos 𝑥
Integrating, we get
∴ 𝑦. sec 𝑥 = sin 𝑥 + 𝑐
∴ 𝑦 = sin 𝑥 cos 𝑥 + 𝑐. cos 𝑥

with Modeling Applications
– Math Tutor at Academic Resource
Center