Lecture 14
Section 15.5 from
Fundamental methods of Mathematical Economics, McGraw Hill 2005, 4th Edition.
By A. C. Chiang & Kevin Wainwright is covered.
Equations reducible to the linear form
If the differential equation
𝑑𝑦
𝑑𝑡
= ℎ(𝑦, 𝑡) happens to take the specific nonlinear form
𝑑𝑦
+ 𝑅(𝑡)𝑦 = 𝑇(𝑡)𝑦 𝑚
(1)
𝑑𝑡
where m is any number other than 0 and 1, then the equation is referred to as a Bernoulli
equation. For m=0 and m=1, equation (1) is linear in y. Bernoulli equation can always be
reduced to a linear ordinary differential equation and be solved as such.
The reduction procedure is relatively simple. First, we can divide (1) by 𝑦 𝑚 , to get
𝑑𝑦
𝑦 −𝑚
+ 𝑅(𝑡)𝑦 1−𝑚 = 𝑇(𝑡) (2)
𝑑𝑡
Introduce 𝑧 = 𝑦1−𝑚 , so that
𝑑𝑧 𝑑𝑧 𝑑𝑦
𝑑𝑦
=
= (1 − 𝑚)𝑦 −𝑚
(3)
𝑑𝑡 𝑑𝑦 𝑑𝑡
𝑑𝑡
Then (2) can be written as
1 𝑑𝑧
+ 𝑅(𝑡)𝑧 = 𝑇(𝑡) (4)
1 − 𝑚 𝑑𝑡
Equation (4) is first-order linear ODE in z. We can find solution z(t) of (3) by calculating
its integrating factor.
Then as a final step we can translate z back to y by the reverse substitution.
Example 1: Solve the ODE
𝑑𝑦
+ 𝑡𝑦 = 3𝑡𝑦 2 (5)
𝑑𝑡
This is Bernoulli’s equation with m=2. Dividing (5) by 𝑦 2 , we have
𝑑𝑦
𝑦 −2
+ 𝑡𝑦 −1 = 3𝑡 (6)
𝑑𝑡
Introduce 𝑧 = 𝑦1−2 = 𝑦 −1 , so that
𝑑𝑧 𝑑𝑧 𝑑𝑦
𝑑𝑦
=
= −𝑦 −2
(7)
𝑑𝑡 𝑑𝑦 𝑑𝑡
𝑑𝑡
Then (6) can be written as
𝑑𝑧
− + 𝑡𝑧 = 3𝑡
𝑑𝑡
𝑜𝑟
𝑑𝑧
− 𝑡𝑧 = −3𝑡 (8)
𝑑𝑡
Equation (8) is linear in z. The integrating factor is
1
Instructor Dr Rehana Naz
Mathematical Economics II
𝑡2
𝐼. 𝐹. = 𝑒 ∫ −𝑡𝑑𝑡 = 𝑒 − 2
𝑡2
Multiplying (8) by 𝑒 − 2 , we have
𝑡2
𝑡2
𝑡2
− 𝑑𝑧
−
−
2
2
𝑒
− 𝑒 𝑡𝑧 = −3𝑡𝑒 2
𝑑𝑡
This yields
𝑡2
𝑡2
𝑑
[𝑧𝑒 − 2 ] = −3𝑡𝑒 − 2
𝑑𝑡
Integrating both sides, we have
𝑡2
𝑡2
𝑑
−
−
2
∫ [𝑧𝑒 ] 𝑑𝑡 = ∫ −3𝑡𝑒 2 𝑑𝑡
𝑑𝑡
𝑡2
𝑡2
𝑧𝑒 − 2 = 3𝑒 − 2 + 𝑐
𝑡2
𝑧(𝑡) = 3 + 𝑐𝑒 2
(9)
Since our primary interest lies in the solution y(t) rather than z(t) , we must perform a
reverse transformation using the equation 𝑧 = 𝑦 −1 , equation (9) takes the following
form
𝑡2
1
= 3 + 𝑐𝑒 2
𝑦
𝑜𝑟
1
𝑦(𝑡) =
𝑡2
3 + 𝑐𝑒 2
as desired solution. This is general solution, because an arbitrary constant c is present.
Note: Equation (8) can also be solved by separation of variables method.
Example 2: Solve the ODE
𝑑𝑦 1
+ 𝑦 = 𝑦 3 (10)
𝑑𝑡 𝑡
This is Bernoulli’s equation with m=3. Dividing (10) by 𝑦 3 , we have
𝑑𝑦 1 −2
𝑦 −3
+ 𝑦 = 1 (11)
𝑑𝑡 𝑡
Introduce 𝑧 = 𝑦1−3 = 𝑦 −2 , so that
𝑑𝑧 𝑑𝑧 𝑑𝑦
𝑑𝑦
=
= −2𝑦 −3
(12)
𝑑𝑡 𝑑𝑦 𝑑𝑡
𝑑𝑡
Then (11) can be written as
1 𝑑𝑧 1
−
+ 𝑧=1
2 𝑑𝑡 𝑡
𝑜𝑟
𝑑𝑧 2
− 𝑧 = −2
(13)
𝑑𝑡 𝑡
2
Instructor Dr Rehana Naz
Mathematical Economics II
Equation (13) is linear in z. The integrating factor is
2
−2
𝐼. 𝐹. = 𝑒 ∫ − 𝑡 𝑑𝑡 = 𝑒 −2𝑙𝑛|𝑡| = 𝑒 𝑙𝑛|𝑡 | = 𝑡 −2
Multiplying (13) by 𝑡 −2, we have
𝑑𝑧
𝑡 −2 − 2𝑡 −3 𝑧 = −2𝑡 −2
𝑑𝑡
This yields
𝑑
[𝑧𝑡 −2 ] = −2𝑡 −2
𝑑𝑡
Integrating both sides, we have
𝑑
∫ [𝑧𝑡 −2 ] 𝑑𝑡 = ∫ −2𝑡 −2 𝑑𝑡
𝑑𝑡
𝑧𝑡 −2 = 2𝑡 −1 + 𝑐
𝑧(𝑡) = 2𝑡 + 𝑐𝑡 2
(14)
Since our primary interest lies in the solution y(t) rather than z(t) , we must perform a
reverse transformation using the equation 𝑧 = 𝑦 −2 , equation (14) takes the following
form
1
= 2𝑡 + 𝑐𝑡 2
𝑦2
𝑜𝑟
1
𝑦(𝑡) =
𝑜𝑟 𝑦(𝑡) = (2𝑡 + 𝑐𝑡 2 ) −1/2
1
(2𝑡 + 𝑐𝑡 2 ) 2
as desired solution. This is general solution, because an arbitrary constant c is present.
Example 3: Solve the ODE
𝑑𝑦
𝑡2
− 2𝑡𝑦 = 3𝑦 4
𝑑𝑡
We can rewrite given ODE as
𝑑𝑦 2
𝑦4
− 𝑦=3 2
(15)
𝑑𝑡 𝑡
𝑡
This is Bernoulli’s equation with m=4. Dividing (15) by 𝑦 4 , we have
𝑑𝑦 2 −3
3
𝑦 −4
− 𝑦 = 2 (16)
𝑑𝑡 𝑡
𝑡
Introduce 𝑧 = 𝑦1−4 = 𝑦 −3 , so that
𝑑𝑧 𝑑𝑧 𝑑𝑦
𝑑𝑦
=
= −3𝑦 −4
(17)
𝑑𝑡 𝑑𝑦 𝑑𝑡
𝑑𝑡
Then (16) can be written as
1 𝑑𝑧 2
3
−
− 𝑧= 2
3 𝑑𝑡 𝑡
𝑡
𝑜𝑟
3
Instructor Dr Rehana Naz
Mathematical Economics II
𝑑𝑧 6
9
+ 𝑧=− 2
(18)
𝑑𝑡 𝑡
𝑡
Equation (18) is linear in z. The integrating factor is
6
6
𝐼. 𝐹. = 𝑒 ∫ 𝑡 𝑑𝑡 = 𝑒 6𝑙𝑛|𝑡| = 𝑒 𝑙𝑛|𝑡 | = 𝑡 6
Multiplying (18) by 𝑡 6 , we have
𝑑𝑧
𝑡6
+ 6𝑡 5 𝑧 = −9𝑡 4
𝑑𝑡
This yields
𝑑
[𝑧𝑡 6 ] = −9𝑡 4
𝑑𝑡
Integrating both sides, we have
𝑑
∫ [𝑧𝑡 6 ] 𝑑𝑡 = ∫ −9𝑡 4 𝑑𝑡
𝑑𝑡
9
𝑧𝑡 6 = − 𝑡 5 + 𝑐
5
9 −1
𝑧(𝑡) = − 𝑡 + 𝑐𝑡 −6
(19)
5
Since our primary interest lies in the solution y(t) rather than z(t) , we must perform a
reverse transformation using the equation 𝑧 = 𝑦 −3 , equation (19) takes the following
form
1
9
= 𝑡 −1 + 𝑐𝑡 −6
3
𝑦
5
𝑜𝑟
1
9
𝑦(𝑡) =
𝑜𝑟 𝑦(𝑡) = (− 𝑡 −1 + 𝑐𝑡 −6 ) −1/3
1
9
5
(− 𝑡 −1 + 𝑐𝑡 −6 ) 3
5
as desired general solution.
Example 4: Solve the ODE
𝑦1/2
𝑑𝑦
+ 𝑦 3/2 = 1
𝑑𝑡
(20)
We can rewrite ODE (20) as
𝑑𝑦
+ 𝑦 = 𝑦 −1/2
(21)
𝑑𝑡
This is Bernoulli’s equation with m=-1/2. Dividing (21) by 𝑦 −1/2, we have
𝑑𝑦
𝑦1/2
+ 𝑦 3/2 = 1 (22)
𝑑𝑡
1
Introduce 𝑧 = 𝑦1−(−2) = 𝑦 3/2 , so that
𝑑𝑧 𝑑𝑧 𝑑𝑦 3 1/2 𝑑𝑦
=
= 𝑦
𝑑𝑡 𝑑𝑦 𝑑𝑡 2
𝑑𝑡
(23)
4
Instructor Dr Rehana Naz
Mathematical Economics II
Then (22) can be written as
2 𝑑𝑧
+𝑧 =1
3 𝑑𝑡
𝑜𝑟
𝑑𝑧 3
3
+ 𝑧=
(24)
𝑑𝑡 2
2
Equation (24) is linear in z. The integrating factor is
3
3
𝐼. 𝐹. = 𝑒 ∫2𝑑𝑡 = 𝑒 2𝑡
3
Multiplying (24) by 𝑒 2𝑡 , we have
3
𝑒 2𝑡
𝑑𝑧 3 3𝑡
3 3
+ 𝑒 2 𝑧 = 𝑒 2𝑡
𝑑𝑡 2
2
This yields
3
𝑑
3 3
[𝑧𝑒 2𝑡 ] = 𝑒 2𝑡
𝑑𝑡
2
Integrating both sides, we have
∫
3
𝑑
3 3
[𝑧𝑒 2𝑡 ] 𝑑𝑡 = ∫ 𝑒 2𝑡 𝑑𝑡
𝑑𝑡
2
3
3
𝑧𝑒 2𝑡 = 𝑒 2𝑡 + 𝑐
3
𝑧(𝑡) = 1 + 𝑐𝑒 −2𝑡
(25)
Since our primary interest lies in the solution y(t) rather than z(t) , we must perform a
reverse transformation using the equation 𝑧 = 𝑦 3/2 , equation (25) takes the following
form
3
𝑦 3/2 = 1 + 𝑐𝑒 −2𝑡
𝑜𝑟
3
𝑦(𝑡) = (1 + 𝑐𝑒 −2𝑡 )
2
3
as desired general solution.
Example 5: Solve the initial value problem
𝑑𝑦
𝑡2
+ 𝑦 2 = 𝑡𝑦, 𝑦(1) = 1/2
(26)
𝑑𝑡
We can rewrite ODE (26) as
𝑑𝑦 𝑦
𝑦2
−
=− 2
(27)
𝑑𝑡
𝑡
𝑡
This is Bernoulli’s equation with m=2. Dividing (27) by 𝑦 2 , we have
𝑑𝑦 𝑦 −1
1
𝑦 −2
−
= − 2 (28)
𝑑𝑡
𝑡
𝑡
Introduce 𝑧 = 𝑦1−2 = 𝑦 −1 , so that
5
Instructor Dr Rehana Naz
Mathematical Economics II
𝑑𝑧 𝑑𝑧 𝑑𝑦
𝑑𝑦
=
= −𝑦 −2
(29)
𝑑𝑡 𝑑𝑦 𝑑𝑡
𝑑𝑡
Then (28) can be written as
𝑑𝑧 𝑧
1
− − =− 2
𝑑𝑡 𝑡
𝑡
𝑜𝑟
𝑑𝑧 𝑧 1
+ =
(30)
𝑑𝑡 𝑡 𝑡 2
Equation (30) is linear in z. The integrating factor is
1
𝐼. 𝐹. = 𝑒 ∫ 𝑡 𝑑𝑡 = 𝑒 𝑙𝑛|𝑡| = 𝑡
Multiplying (30) by 𝑡, we have
𝑑𝑧
1
𝑡 +𝑧 =
𝑑𝑡
𝑡
This yields
𝑑
1
[𝑧𝑡] =
𝑑𝑡
𝑡
Integrating both sides, we have
𝑑
1
∫ [𝑧𝑡] 𝑑𝑡 = ∫ 𝑑𝑡
𝑑𝑡
𝑡
𝑧𝑡 = 𝑙𝑛|𝑡| + 𝑐
𝑙𝑛|𝑡| + 𝑐
𝑧(𝑡) =
(31)
𝑡
Since our primary interest lies in the solution y(t) rather than z(t) , we must perform a
reverse transformation using the equation 𝑧 = 𝑦 −1 , equation (31) takes the following
form
𝑙𝑛|𝑡| + 𝑐
𝑦 −1 =
𝑡
𝑜𝑟
𝑡
𝑦(𝑡) =
(32)
𝑙𝑛|𝑡| + 𝑐
as general solution. Using 𝑦(1) = 1/2, we have
1
1
=
𝑙𝑛|1| + 𝑐 2
Or
1 1
= 𝑜𝑟 𝑐 = 2
𝑐 2
The definite solution is
𝑦(𝑡) =
𝑡
𝑙𝑛|𝑡| + 2
6
Instructor Dr Rehana Naz
Mathematical Economics II
Exercises: Solve following Bernoulli equations
(i)
(ii)
𝑦
2𝑡
𝑦+𝑡
𝑑𝑦
𝑑𝑡
𝑑𝑦
𝑑𝑦 + 𝑦+𝑡 𝑑𝑡 = 0
𝑡
= −𝑦
(iii) 𝑑𝑡 = 3𝑦 2 𝑡
Summary:
(a) First-order linear ordinary differential equation
The general form of first-order linear ordinary differential equation is
𝑑𝑦
+ 𝑢(𝑡)𝑦 = 𝑤(𝑡)
𝑑𝑡
(𝐴)
Where 𝑢(𝑡)and 𝑤(𝑡) can be constants or any functions of t.
There are three methods to solve ODE of type (A).
i.
Separable variables
ii.
Homogeneous ordinary differential equations
iii.
Exact ordinary differential equations
iv.
Solution by integrating factor method
(b) First-order and first-degree nonlinear ordinary differential equation
The general form of first-order linear ordinary differential equation is
𝑑𝑦
= ℎ(𝑦, 𝑡)
𝑑𝑡
(𝐵)
where ℎ(𝑦, 𝑡) is any function of y and t.
There are three methods to solve ODE of type (B).
i. Separable variables
ii. Homogeneous ordinary differential equations
iii. Exact ordinary differential equations, Solution by integrating factor method for
inexact ordinary differential equations
iv. By reducing to the linear form (Bernoulli equations)
7
Instructor Dr Rehana Naz
Mathematical Economics II