1. Differential Equations
Lecture 6 Solution of Second Order Differential Equations Homogeneous Differential Equations fall semester
Instructor: A. S. Brwa / MSc. In Structural Engineering
College of Engineering / Ishik University

2. 2nd Order Linear Homogenous ODE
Ishik University
2nd Order Linear Homogenous ODE
𝑨 𝑟 2 +𝑩𝑟+𝑪=0 Notice that it is an algebraic equation that is obtained from the differential equation by replacing 𝑦′′ by 𝑟 2 , 𝑦 ′ by 𝑟, and 𝑦 by 1. Sometimes the roots 𝑟 1 and 𝑟 2 of this auxiliary equation can be found by factoring. In other cases they are found by using the quadratic formula: 𝒓= −𝒃± 𝒃 𝟐 −𝟒𝒂𝒄 𝟐𝒂
Faculty of Engineering – Differential Equations – Lecture 5 – Second Order Homogeneous DE

3. 2nd Order Linear Homogenous ODE
Ishik University
2nd Order Linear Homogenous ODE
𝑨 𝑟 2 +𝑩𝑟+𝑪=0
𝒓= −𝒃± 𝒃 𝟐 −𝟒𝒂𝒄 𝟐𝒂
We distinguish three cases according to the sign of the discriminant 𝒃 𝟐 −𝟒𝒂𝒄
Case 1: 𝒃 𝟐 −𝟒𝒂𝒄>𝟎 There are two distinct real roots 𝑟 1 and 𝑟 2
Case 2: 𝒃 𝟐 −𝟒𝒂𝒄=𝟎 There is one repeated real root r
Case 3: 𝒃 𝟐 −𝟒𝒂𝒄<𝟎 There are two complex conjugate roots
𝑟=𝛽±𝜇𝑖
Faculty of Engineering – Differential Equations – Lecture 6 – Solution of Second Order Homogeneous ODE

4. 2nd Order Linear Homogenous ODE
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2nd Order Linear Homogenous ODE
Case 2: 𝒃 𝟐 −𝟒𝒂𝒄=𝟎
The roots of the auxiliary equation are real and equal. Let’s denote by the common value of 𝑟 1 and 𝑟 2
So, 2a𝑟+𝑏=0
𝒓 𝟏 equal 𝒓 𝟐
𝒓= −𝒃± 𝒃 𝟐 −𝟒𝒂𝒄 𝟐𝒂
𝒓= −𝒃 𝟐𝒂
becomes
Faculty of Engineering – Differential Equations – Lecture 6 – Solution of Second Order Homogeneous ODE

5. 2nd Order Linear Homogenous ODE
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2nd Order Linear Homogenous ODE
In lecture 5 we have proofed that 𝑦 1 = 𝑒 𝑟 1 𝑥 is a solution of this 2nd order ODE 𝑨 𝒚 ′′ +𝑩 𝒚 ′ +𝑪𝒚=𝟎 We now verify that 𝑦 2 =𝑥 𝑒 𝑟 2 𝑥 is also a solution: 𝒚 ′ = 𝒆 𝒓 𝟐 𝒙 + 𝑟 2 𝒙𝒆 𝒓 𝟐 𝒙 𝒚 ′′ =𝟐 𝒆 𝒓 𝟐 𝒙 + 𝒓 𝟐 𝒙 𝒆 𝒓 𝟐 𝒙
Faculty of Engineering – Differential Equations – Lecture 6 – Solution of Second Order Homogeneous ODE

6. 2nd Order Linear Homogenous ODE
Ishik University
2nd Order Linear Homogenous ODE
Write down the previous equation in terms of 𝑦 2 and substitute the values of 𝒚 𝟐 ′ and 𝒚 𝟐 ′′ we get 𝑨 𝒚 𝟐 ′′ +𝑩 𝒚 𝟐 ′ +𝑪 𝒚 𝟐 =𝟎 𝑨 𝟐 𝒆 𝒓 𝟐 𝒙 + 𝒓 𝟐 𝒙 𝒆 𝒓 𝟐 𝒙 +𝑩 𝒆 𝒓 𝟐 𝒙 + 𝑟 2 𝒙𝒆 𝒓 𝟐 𝒙 +𝑪 𝒙 𝒆 𝒓 𝟐 𝒙 =𝟎 𝑨𝟐 𝒆 𝒓 𝟐 𝒙 +𝑨 𝒓 𝟐 𝒙 𝒆 𝒓 𝟐 𝒙 +𝑩 𝒆 𝒓 𝟐 𝒙 +𝑩 𝑟 2 𝒙𝒆 𝒓 𝟐 𝒙 +𝑪 𝒙 𝒆 𝒓 𝟐 𝒙 =𝟎 𝟐𝒓𝑨+𝑩 𝒆 𝒓 𝟐 𝒙 + 𝑨 𝑟 2 +𝑩𝑟+𝑪 𝒙 𝒆 𝒓 𝟐 𝒙 =𝟎 (𝟎) 𝒆 𝒓𝒙 +(𝟎) 𝒙𝒆 𝒓𝒙 =𝟎
Faculty of Engineering – Differential Equations – Lecture 6 – Solution of Second Order Homogeneous ODE

7. 2nd Order Linear Homogenous ODE
Ishik University
2nd Order Linear Homogenous ODE
If the auxiliary equation 𝑨 𝑟 2 +𝑩𝑟+𝑪 has only one real root r, then the general solution of 𝑨 𝒚 𝟐 ′′ +𝑩 𝒚 𝟐 ′ +𝑪 𝒚 𝟐 =𝟎 is
𝒚= 𝒄 𝟏 𝒆 𝒓 𝟏 𝒙 + 𝒄 𝟐 𝒙𝒆 𝒓 𝟐 𝒙
Since 𝒓 𝟏 = 𝒓 𝟐
𝒚= 𝒄 𝟏 𝒆 𝒓𝒙 + 𝒄 𝟐 𝒙𝒆 𝒓𝒙
Faculty of Engineering – Differential Equations – Lecture 6 – Solution of Second Order Homogeneous ODE

8. 2nd Order Linear Homogenous ODE
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2nd Order Linear Homogenous ODE
Example 1: 4 𝒚 ′′ +𝟏𝟐 𝒚 ′ +𝟗𝒚=𝟎
SOLUTION: The auxiliary equation 𝟒 𝑟 2 +12𝑟+𝟗=𝟎 can be factored as
2𝑟+3 2𝑟+3 =0
so the only root is 𝑟=− 3 2
𝒚= 𝒄 𝟏 𝒆 − 3 2 𝒙 + 𝒄 𝟐 𝒙𝒆 − 3 2 𝒙
Faculty of Engineering – Differential Equations – Lecture 6 – Solution of Second Order Homogeneous ODE

9. 2nd Order Linear Homogenous ODE
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2nd Order Linear Homogenous ODE
Example 2: 𝒚 ′′ −𝟒 𝒚 ′ +𝟒𝒚=𝟎
SOLUTION: The auxiliary equation 𝑟 2 −𝟒𝑟+𝟒=𝟎 can be factored as
𝑟−2 𝑟−2 =0
so the only root is 𝑟=2
𝒚= 𝒄 𝟏 𝒆 2𝒙 + 𝒄 𝟐 𝒙𝒆 2𝒙
Faculty of Engineering – Differential Equations – Lecture 6 – Solution of Second Order Homogeneous ODE

10. 2nd Order Linear Homogenous ODE
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2nd Order Linear Homogenous ODE
For example 2 If initial condition is given 𝒚 𝟎 =𝟒 and 𝒚 ′ (𝟎)=𝟓 𝒚 ′′ −𝟒 𝒚 ′ +𝟒𝒚=𝟎 SOLUTION:
𝒚= 𝒄 𝟏 𝒆 2𝒙 + 𝒄 𝟐 𝒙𝒆 2𝒙
𝟒= 𝒄 𝟏 𝒆 2(𝟎) + 𝒄 𝟐 (𝟎)𝒆 2(𝟎)
𝟒= 𝒄 𝟏 𝟏
𝒄 𝟏 =𝟒
𝒚 ′ =𝟐 𝒄 𝟏 𝒆 2𝒙 + 𝒄 𝟐 𝟐𝒙𝒆 2𝒙 + 𝒆 𝟐𝒙
𝟓=𝟐 𝒄 𝟏 𝒆 2(𝟎) + 𝒄 𝟐 ( 𝟐(𝟎)𝒆 2(𝟎) + 𝒆 𝟐(𝟎) )
𝟓=𝟐 𝒄 𝟏 𝟏+ 𝒄 𝟐 (𝟎+𝟏)
𝟓=𝟐(𝟒)𝟏+ 𝒄 𝟐 (𝟎+𝟏)
𝒄 𝟐 =𝟓−𝟖=−𝟑
Faculty of Engineering – Differential Equations – Lecture 6 – Solution of Second Order Homogeneous ODE

11. 2nd Order Linear Homogenous ODE
Ishik University
2nd Order Linear Homogenous ODE
Case 3: 𝒃 𝟐 −𝟒𝒂𝒄<𝟎
the characteristic polynomial has two complex roots, which are conjugates, 𝒓 𝟏 =𝜷+𝝁𝒊 and 𝒓 𝟐 =𝜷−𝝁𝒊
(λ, µ are real numbers, µ > 0). As before they give two linearly independent solutions 𝑦 1 = 𝑒 𝑟 1 𝑥 and 𝑦 2 = 𝑒 𝑟 2 𝑥 Consequently the linear combination will be a general solution.
𝒚= 𝒄 𝟏 𝒆 𝒓 𝟏 𝒙 + 𝒄 𝟐 𝒆 𝒓 𝟐 𝒙
𝑟=𝛽±𝜇𝑖
Faculty of Engineering – Differential Equations – Lecture 6 – Solution of Second Order Homogeneous ODE

12. 2nd Order Linear Homogenous ODE
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2nd Order Linear Homogenous ODE
At this juncture you might have this question: “but aren’t 𝑟 1 and 𝑟 2 complex numbers; what would become of the exponential function with a complex number exponent?” The answer to that question is given by the Euler’s formula. Euler’s formula, for any real number θ,
𝒆 𝜽 𝒊 = 𝒄𝒐𝒔 𝜽+𝒊 𝒔𝒊𝒏 𝜽
Faculty of Engineering – Differential Equations – Lecture 6 – Solution of Second Order Homogeneous ODE

13. 2nd Order Linear Homogenous ODE
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2nd Order Linear Homogenous ODE
Hence, when r is a complex number λ + µi, the exponential function 𝒆 𝒓𝒙 becomes
𝒆 𝒓𝒙 = 𝒆 (λ + µi)𝒙 = 𝒆 λ𝒙 𝒆 µi𝒙 = 𝒆 λ𝒙 cos µ𝒙 +i sin µ𝒙
Similarly, when r = λ − µi , 𝒆 𝒓𝒙 becomes
𝒆 𝒓𝒙 = 𝒆 (λ − µi)𝒙 = 𝒆 λ𝒙 𝒆 −µi𝒙 = 𝒆 λ𝒙 cos(− µ𝒙) +i sin (−µ𝒙)
= 𝒆 λ𝒙 cos µ𝒙 −i sin µ𝒙
Faculty of Engineering – Differential Equations – Lecture 6 – Solution of Second Order Homogeneous ODE

14. 2nd Order Linear Homogenous ODE
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2nd Order Linear Homogenous ODE
Hence, the general solution found above is then
𝒚= 𝒄 𝟏 𝒆 λ𝒙 cos µ𝒙 +i sin µ𝒙 + 𝒄 𝟐 𝒆 λ𝒙 cos µ𝒙 −i sin µ𝒙
However, this general solution is a complex-valued function (meaning that, given a real number t, the value of the function y(t) could be complex). It represents the general form of all particular solutions with either real or complex number coefficients. What we seek here, instead, is a real-valued expression that gives only the set of all particular solutions with real number coefficients only.
Faculty of Engineering – Differential Equations – Lecture 6 – Solution of Second Order Homogeneous ODE

15. 2nd Order Linear Homogenous ODE
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2nd Order Linear Homogenous ODE
Now, we keep only those whose coefficients are real numbers.
𝒖(𝒙)= 𝒆 λ𝒙 cos µ𝒙
𝒗(𝒙)= 𝒆 λ𝒙 sin µ𝒙
It is easy to verify that both u and v satisfy the differential equation (one way to see this is to observe that u can be obtain from the complex-valued general solution by setting 𝒄 𝟏 = 𝒄 𝟐 =𝟏/𝟐
𝒖(𝒙)= 𝟏 𝟐 𝒆 λ𝒙 cos µ𝒙 +i sin µ𝒙 + 𝟏 𝟐 𝒆 λ𝒙 cos µ𝒙 −i sin µ𝒙
𝒖(𝒙)= 𝟏 𝟐 𝒆 λ𝒙 cos µ𝒙 + 𝟏 𝟐 𝒆 λ𝒙 i sin µ𝒙 + 𝟏 𝟐 𝒆 λ𝒙 cos µ𝒙 − 𝟏 𝟐 𝒆 λ𝒙 i sin µ𝒙
Faculty of Engineering – Differential Equations – Lecture 6 – Solution of Second Order Homogeneous ODE

16. 2nd Order Linear Homogenous ODE
Ishik University
2nd Order Linear Homogenous ODE
Now, we keep only those whose coefficients are real numbers.
𝒖(𝒙)= 𝒆 λ𝒙 cos µ𝒙
𝒗(𝒙)= 𝒆 λ𝒙 sin µ𝒙
It is easy to verify that both u and v satisfy the differential equation (one way to see this is to observe that v can be obtain from the complex-valued general solution by setting 𝒄 𝟏 = 𝟏 𝟐𝒊 𝑎𝑛𝑑 𝒄 𝟐 =− 𝟏 𝟐𝒊
𝒗(𝒙)= 𝟏 𝟐𝒊 𝒆 λ𝒙 cos µ𝒙 +i sin µ𝒙 + − 𝟏 𝟐𝒊 𝒆 λ𝒙 cos µ𝒙 −i sin µ𝒙
𝒗(𝒙)= 𝟏 𝟐𝒊 𝒆 λ𝒙 cos µ𝒙 + 𝟏 𝟐𝒊 𝒆 λ𝒙 i sin µ𝒙 + − 𝟏 𝟐𝒊 𝒆 λ𝒙 cos µ𝒙 − − 𝟏 𝟐𝒊 𝒆 λ𝒙 i sin µ𝒙
Faculty of Engineering – Differential Equations – Lecture 6 – Solution of Second Order Homogeneous ODE

17. 2nd Order Linear Homogenous ODE
Ishik University
2nd Order Linear Homogenous ODE
Therefore, the functions u and v are linearly independent solutions of the equation. They form a pair of real-valued fundamental solutions and the linear combination is a desired real-valued general solution:
𝒚= 𝒄 𝟏 𝒆 λ𝒙 cos µ𝒙 + 𝒄 𝟐 𝒆 λ𝒙 sin µ𝒙
Faculty of Engineering – Differential Equations – Lecture 6 – Solution of Second Order Homogeneous ODE

18. 2nd Order Linear Homogenous ODE
Ishik University
2nd Order Linear Homogenous ODE
Example 3: 𝒚 ′′ −𝟔 𝒚 ′ +𝟏𝟑𝒚=𝟎 SOLUTION: The auxiliary equation 𝒓 𝟐 −𝟔𝒓+𝟏𝟑=𝟎 cant be factored. So use quadratic formula 𝒓= −(−𝟔)± −𝟏𝟔 𝟐 and
𝒓= ± −𝟏𝟔 𝟒 =±𝟐𝒊
𝒓= 𝟔 𝟐 =𝟑
𝒓=𝟑±𝟐𝒊
Faculty of Engineering – Differential Equations – Lecture 6 – Solution of Second Order Homogeneous ODE

19. 2nd Order Linear Homogenous ODE
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2nd Order Linear Homogenous ODE
Example 3: 𝒚 ′′ −𝟔 𝒚 ′ +𝟏𝟑𝒚=𝟎
SOLUTION: The auxiliary equation 𝒓 𝟐 −𝟔𝒓+𝟏𝟑=𝟎 cant be factored. So use quadratic formula
𝒓= −(−𝟔)± −𝟏𝟔 𝟐
𝒚= 𝒆 3𝒙 𝒄 𝟏 cos 2𝒙 + 𝒄 𝟐 sin 2𝒙
𝒓=𝟑±𝟐𝒊
Faculty of Engineering – Differential Equations – Lecture 6 – Solution of Second Order Homogeneous ODE