Ordinary Differential Equations I
Lecture (5)
First Order Differential Equations
7. Riccati Equations
The equation
𝑑𝑦
+ 𝑝(𝑥)𝑦 + 𝑄(𝑥)𝑦 2 = 𝐹(𝑥)
𝑑𝑥
Is known as Riccati equation.
Use of Particular Solutions to Construct the General Solution
Suppose that some particular solution 𝑦1 of this equation is known.
𝑑𝑦1
⇒ + 𝑝(𝑥)𝑦1 + 𝑄(𝑥)(𝑦1 )2 = 𝐹(𝑥)
𝑑𝑥
A more general solution containing one arbitrary constant can be obtained through the
substitution
𝑑𝑦 𝑑𝑦1 𝑑𝑣
𝑦 = 𝑦1 (𝑥) + 𝑣(𝑥) ⇒ = +
𝑑𝑥 𝑑𝑥 𝑑𝑥
So, we have
𝑑𝑦1 𝑑𝑣
[ + ] + 𝑝(𝑥)[𝑦1 (𝑥) + 𝑣(𝑥)] + 𝑄(𝑥)[𝑦1 (𝑥) + 𝑣(𝑥)]2 = 𝐹(𝑥)
𝑑𝑥 𝑑𝑥
𝑑𝑦1 𝑑𝑣
[ + 𝑝(𝑥)𝑦1 + 𝑄(𝑥)(𝑦1 )2 ] + + [𝑝(𝑥) + 𝑦1 (𝑥)𝑄(𝑥)]𝑣 + 𝑄(𝑥)𝑣 2 = 𝐹(𝑥)
𝑑𝑥 𝑑𝑥
𝑑𝑣
𝐹(𝑥) + + [𝑝(𝑥) + 𝑦1 (𝑥)𝑄(𝑥)]𝑣 + 𝑄(𝑥)𝑣 2 = 𝐹(𝑥)
𝑑𝑥
𝑑𝑣
+ [𝑝(𝑥) + 𝑦1 (𝑥)𝑄(𝑥)]𝑣 + 𝑄(𝑥)𝑣 2 = 0 is Bernoulli equation.
𝑑𝑥
Note:
𝑑𝑦
1. If 𝑄(𝑥) = 0, The Riccati equation reduces to a linear first order equation (𝑑𝑥 + 𝑝(𝑥)𝑦 = 𝐹(𝑥)).
𝑑𝑦
2. If 𝐹(𝑥) = 0, The Riccati equation reduces to a Bernoulli equation (𝑑𝑥 + 𝑝(𝑥)𝑦 + 𝑄(𝑥)𝑦 2 = 0).
Lecture: Hakima Kh. Ahmed 1
Example (1): Solve 𝑥 3 𝑦 ′ + 𝑥 2 𝑦 − 𝑦 2 = 2𝑥 4 , given that 𝑦1 (𝑥) = 2𝑥 2
Solution:
The given differential equation can be written as
𝑑𝑦 1 1
+ 𝑦− 𝑦 2 = 2𝑥 which is a Riccati equation
𝑑𝑥 𝑥 𝑥3
𝑑𝑦 𝑑𝑣
let 𝑦 = 𝑦1 (𝑥) + 𝑣(𝑥) = 2𝑥 2 + 𝑣(𝑥) ⇒ = 4𝑥 +
𝑑𝑥 𝑑𝑥
𝑑𝑦
Sub y and , in the Riccati equation we get
𝑑𝑥
𝑑𝑣 1 1
(4𝑥 + 𝑑𝑥) + 𝑥 (2𝑥 2 + 𝑣(𝑥)) − 𝑥 3 (2𝑥 2 + 𝑣(𝑥))2 = 2𝑥
𝑑𝑣 1 4 1
4𝑥 + + 2𝑥 + 𝑣(𝑥) − 4𝑥 − 𝑣(𝑥) − 𝑣(𝑥)2 = 2𝑥
𝑑𝑥 𝑥 𝑥 𝑥3
𝑑𝑣 3 1
− 𝑣(𝑥) − 𝑣(𝑥)2 = 0
𝑑𝑥 𝑥 𝑥3
𝑑𝑣 3 1
− 𝑣(𝑥) = 𝑣(𝑥)2 is a Bernoulli equation
𝑑𝑥 𝑥 𝑥3
𝑑𝑣 3 1
𝑣(𝑥)−2 − 𝑣(𝑥)−1 =
𝑑𝑥 𝑥 𝑥3
𝑑𝑢 𝑑𝑣
Let 𝑢 = 𝑣(𝑥)−1 ⇒ − = 𝑣(𝑥)−2
𝑑𝑥 𝑑𝑥
𝑑𝑢 3 −1
⇒ + 𝑢= is a linear equation
𝑑𝑥 𝑥 𝑥3
3 −1
𝑝(𝑥) = , 𝑄(𝑥) =
𝑥 𝑥3
3
∫𝑥𝑑𝑥
𝐼(𝑥) = 𝑒 ∫ 𝑝(𝑥)𝑑𝑥 =𝑒 = 𝑥3
1
𝑢= [∫ 𝐼(𝑥)𝑄(𝑥) dx]
𝐼(𝑥)
1 1
𝑢= [− ∫ 𝑥 3 dx]
𝑥3 𝑥3
1
𝑢= [−𝑥 + 𝑐] ∵ 𝑢 = 𝑣(𝑥)−1
𝑥3
−𝑥+𝑐
∴ 𝑣(𝑥)−1 = ∵ 𝑦 = 2𝑥 2 + 𝑣(𝑥) ⟹ 𝑣(𝑥) = 𝑦 − 2𝑥 2
𝑥3
The general solution is
𝑥3
∴ 𝑦 − 2𝑥 2 =
−𝑥+𝑐
Lecture: Hakima Kh. Ahmed 2
𝑑𝑦 1 1 1
Example (2): Solve + 𝑦 − 𝑦2 = − , when y = 𝑖𝑠 𝑎 𝑝𝑎𝑟𝑡𝑖𝑐𝑢𝑙𝑎𝑟 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 .
𝑑𝑥 𝑥 𝑥2 𝑥
Solution:
1 𝑑𝑦 −1 𝑑𝑣
let 𝑦 = 𝑦1 (𝑥) + 𝑣(𝑥) = + 𝑣(𝑥) ⇒ = +
𝑥 𝑑𝑥 𝑥2 𝑑𝑥
𝑑𝑦
Sub y and , in the given D.E., we get
𝑑𝑥
2
−1 𝑑𝑣 1 1 1 1
( 𝑥 2 + 𝑑𝑥) + 𝑥 (𝑥 + 𝑣(𝑥)) − (𝑥 + 𝑣(𝑥)) = − 𝑥 2
−1 𝑑𝑣 1 1 1 2 1
+ + + 𝑣(𝑥) − − 𝑣(𝑥) − 𝑣(𝑥)2 = −
𝑥2 𝑑𝑥 𝑥2 𝑥 𝑥2 𝑥 𝑥2
𝑑𝑣 1
− 𝑣(𝑥) − 𝑣(𝑥)2 = 0
𝑑𝑥 𝑥
𝑑𝑣 1
− 𝑣(𝑥) = 𝑣(𝑥)2 is a Bernoulli equation
𝑑𝑥 𝑥
𝑑𝑣 1
𝑣(𝑥)−2 − 𝑣(𝑥)−1 = 1
𝑑𝑥 𝑥
𝑑𝑢 𝑑𝑣
Let 𝑢 = 𝑣(𝑥)−1 ⇒ − = 𝑣(𝑥)−2
𝑑𝑥 𝑑𝑥
𝑑𝑢 1
⇒ + 𝑢 = −1 is a linear equation
𝑑𝑥 𝑥
1
𝑝(𝑥) = , 𝑄(𝑥) = −1
𝑥
1
𝐼(𝑥) = 𝑒 ∫ 𝑝(𝑥)𝑑𝑥 = 𝑒 ∫𝑥𝑑𝑥 = 𝑒 ln(𝑥) = 𝑥
1
𝑢= [∫ 𝐼(𝑥)𝑄(𝑥) dx]
𝐼(𝑥)
1
𝑢 = [− ∫ 𝑥 dx]
𝑥
1 −𝑥 2
𝑢= [ + 𝑐]
𝑥 2
−𝑥 2 +2𝑐
𝑣(𝑥)−1 =
2𝑥
1 2𝑥
𝑦− =
𝑥 −𝑥 2 +2𝑐
8. Linear Substitutions
A differential equation of the form
𝑑𝑦
= 𝑓(𝑎𝑥 + 𝑏𝑦 + 𝑐)
𝑑𝑥
Lecture: Hakima Kh. Ahmed 3
can always be reduced to an equation with separable variables by means of the substitution 𝑢 =
𝑎𝑥 + 𝑏𝑦,𝑏 ≠ 0.
𝑑𝑦 1
Example (1): Solve =
𝑑𝑥 2𝑥−4𝑦+7
Solution:
Let 𝑢 = 2𝑥 − 4𝑦 + 7
1
⟹ 𝑦 = (2𝑥 − 𝑢 + 7)
4
𝑑𝑦 1 𝑑𝑢
𝑎𝑛𝑑 = (2 − )
𝑑𝑥 4 𝑑𝑥
After the substitution, the original DE becomes
1 𝑑𝑢 1
4
(2 − 𝑑𝑥 ) = 𝑢
𝑑𝑢 4
2− =
𝑑𝑥 𝑢
𝑑𝑢 4
=2−
𝑑𝑥 𝑢
𝑑𝑢 2𝑢−4 2(𝑢−2)
= =
𝑑𝑥 𝑢 𝑢
𝑢
(𝑢−2)
𝑑𝑢 = 2𝑑𝑥
we can integrate both sides of our last differential equation for u,
𝑢
∫ (𝑢−2) 𝑑𝑢 = ∫ 2𝑑𝑥
noticing that
𝑢 𝑢−2+2 2
(𝑢−2)
= =1+
𝑢−2 𝑢−2
2
⟹ ∫ (1 + ) 𝑑𝑢 = ∫ 2𝑑𝑥
𝑢−2
⟹ 𝑢 + 2 ln(𝑢 − 2) = 2𝑥 + 𝑐 ∵ 𝑢 = 2𝑥 − 4𝑦 + 7
The general solution is
2𝑥 − 4𝑦 + 7 + 2 ln((2𝑥 − 4𝑦 + 7) − 2) = 2𝑥 + 𝑐
2 ln(2𝑥 − 4𝑦 + 5) − 4𝑦 + 7 = 2𝑥 + 𝑐
Lecture: Hakima Kh. Ahmed 4
Example (2): Find the general solution of the following D.E.
𝑦 ′ = (𝑥 + 𝑦 + 3)2
Solution:
Let 𝑢 = 𝑥 + 𝑦 + 3
⟹𝑦 =𝑢−𝑥−3
𝑑𝑢
and 𝑦 ′ = −1
𝑑𝑥
After the substitution, the original DE becomes
𝑑𝑢
− 1 = 𝑢2
𝑑𝑥
𝑑𝑢
= 𝑢2 + 1
𝑑𝑥
1 𝑑𝑢
= 𝑑𝑥
𝑢2 + 1 𝑑𝑥
we can integrate both sides of our last differential equation for u,
1 𝑑𝑢
∫ = ∫ 𝑑𝑥
𝑢2 + 1 𝑑𝑥
𝑡𝑎𝑛−1 (𝑢) = 𝑥 + 𝑐 ∵u=𝑥+𝑦+3
The general solution is
𝑡𝑎𝑛−1 (𝑥 + 𝑦 + 3) = 𝑥 + 𝑐
Exercise:
Q (1) The D.E.
𝑑𝑦
= 𝐹(𝑥) + 𝑝(𝑥)𝑦 + 𝑄(𝑥)𝑦 2
𝑑𝑥
is called a Riccati D.E. Suppose that one Particular solution 𝑦1 (𝑥) of this D.E. is known, show
1
that the substitution 𝑦 = 𝑦1 (𝑥) + can transform the Riccati D.E. into a linear D.E.
𝑣(𝑥)
𝑑𝑣
+ (𝑝(𝑥) + 2𝑄(𝑥)𝑦1 (𝑥))𝑣 = −𝑄(𝑥)
𝑑𝑥
Q (2) solve the following differential equations
𝑑𝑦 2𝑐𝑜𝑠 2 (𝑥)−𝑠𝑖𝑛2 (𝑥)+𝑦 2
1) = , given that 𝑦1 (𝑥) = 𝑠𝑖𝑛(𝑥)
𝑑𝑥 2𝑐𝑜𝑠(𝑥)
Lecture: Hakima Kh. Ahmed 5
1
2) 𝑦 ′ = 1 + (𝑥 2 − 2𝑥𝑦 + 𝑦 2 ) , given that 𝑦1 (𝑥) = 𝑥
4
3) 𝑦 = 1 + 𝑥 2 − 2𝑥𝑦 + 𝑦 2
′
4) (𝑥 + 𝑦)𝑦 ′ = 𝑦
5) 𝑦 ′ + 𝑦 2 = 1 + 𝑥 2 , given that 𝑦1 (𝑥) = 𝑥
2 1
6) 𝑦 ′ + 𝑦 2 = , given that 𝑦1 (𝑥) = −
𝑥2 𝑥
′
7) 𝑦 − 2√2𝑥 + 𝑦 − 3 = −2
1
8) 𝑦 ′ = 4 +
𝑠𝑖𝑛(4𝑥−𝑦)
𝑑𝑦 (3𝑥−2)2 +1 3
Q (3) Using a linear substitution, solve = + .
𝑑𝑥 3𝑥−2𝑦 2
References
1. Dennis G. Zill, (2016), A First Course in Differential Equations with Modeling
Application (11𝑡ℎ Edition).
2. William E. Boyce, Richard C. Diprima, Douglas B. Meade, (2017) Elementary
Differential Equations and Boundary Value Problems (11𝑡ℎ Edition)
Lecture: Hakima Kh. Ahmed 6
