1
Md. Al-Amin
ID: 172015031
Department: CSE
2
Problem & Solution
(Higher Order Differential Equations)
From-
Ordinary & Partial Differential Equations
(academic purpose book)
Course ode- (MATH-103)
3
Higher Order Differential Equations
Higher Differential Equation-
A higher order differential equation is an equation containing
dependent and independent variables and two or more derivatives
of the dependent variable with respect to one or more independent
variable.
Solve Hints:
If the Roots are Real and Equal – ( 𝑐1 + 𝑐2^x)𝑒 𝑚𝑥
If the Roots are Real and Unequal - ( 𝑐1 𝑒 𝑚1 𝑥
+ 𝑐2 𝑒 𝑚2 𝑥
)
If the Roots are Complex conjugate - 𝑒 𝑎𝑥[𝐴 cos 𝑏𝑥 + 𝐵 sin 𝑏𝑥] 4
𝑑2 𝑦
𝑑𝑥2 − 3
𝑑𝑦
𝑑𝑥
+ 2𝑦 = 0, x=0, y=2,
𝑑𝑦
𝑑𝑥
= 0
Problem
5
𝒅 𝟐 𝒚
𝒅𝒙 𝟐
− 𝟑
𝒅𝒚
𝒅𝒙
+ 𝟐𝒚 = 𝟎 −− −(𝟏)
Let, y = 𝑒 𝑚𝑥
be the trail solution.
∴
𝑑𝑦
𝑑𝑥
= 𝑚𝑒 𝑚𝑥 (diff.w.r.to x)
∴
𝑑2 𝑦
𝑑𝑥2 = 𝑚2 𝑒 𝑚𝑥 (again diff.w.r.to x)
From equation (1)-
𝑚2 𝑒 𝑚𝑥 - 3𝑚𝑒 𝑚𝑥 + 2𝑚𝑒 𝑚𝑥 =0
⇒ 𝑒 𝑚𝑥 (𝑚2 − 3𝑚 + 2) = 0
⇒ 𝑚2 − 3𝑚 + 2 = 0
⇒ 𝑚2
− 2𝑚 − 𝑚 + 2 = 0
⇒ m (m -2) -1 (m -2) = 0
⇒ (m -2) (m-1) = 0
∴ m = 1, 2
Solution
6
Roots are real and unequal.
So the general solution is-
y = 𝑐1 𝑒 𝑚1 𝑥 + 𝑐2 𝑒 𝑚2 𝑥
∴ y = 𝑐1 𝑒 𝑥
+ 𝑐2 𝑒2𝑥
−− −(2)
⇒ 2 = 𝑐1 𝑒0 + 𝑐2 𝑒0 (∵ 𝑥 = 0)
⇒ 2 = 𝑐1 + 𝑐2 −− − 3
Now from equation (2) –
𝑑𝑦
𝑑𝑥
= 𝑐1 𝑒 𝑥 + 2𝑐2 𝑒2𝑥
⇒ 0 = 𝑐1 𝑒0
+ 2𝑐2 𝑒0
⇒ 0 = 𝑐1 + 2𝑐2 − − − (4)
Now (3) – (4)
2 = 𝑐1 + 𝑐2
0 = 𝑐1 + 2𝑐2
2 = − 𝑐2
⇒ 𝑐2 = −2
From equation (3)
2 = 𝑐1 + −2
⇒ − 𝑐1= − 2 − 2
⇒ 𝑐1= 4
Hence the general
solution is –
y = 4𝑒 𝑥 - 2𝑒2𝑥 (Ans:) 7
8

Math presentation

  • 1.
  • 2.
  • 3.
    Problem & Solution (HigherOrder Differential Equations) From- Ordinary & Partial Differential Equations (academic purpose book) Course ode- (MATH-103) 3
  • 4.
    Higher Order DifferentialEquations Higher Differential Equation- A higher order differential equation is an equation containing dependent and independent variables and two or more derivatives of the dependent variable with respect to one or more independent variable. Solve Hints: If the Roots are Real and Equal – ( 𝑐1 + 𝑐2^x)𝑒 𝑚𝑥 If the Roots are Real and Unequal - ( 𝑐1 𝑒 𝑚1 𝑥 + 𝑐2 𝑒 𝑚2 𝑥 ) If the Roots are Complex conjugate - 𝑒 𝑎𝑥[𝐴 cos 𝑏𝑥 + 𝐵 sin 𝑏𝑥] 4
  • 5.
    𝑑2 𝑦 𝑑𝑥2 −3 𝑑𝑦 𝑑𝑥 + 2𝑦 = 0, x=0, y=2, 𝑑𝑦 𝑑𝑥 = 0 Problem 5
  • 6.
    𝒅 𝟐 𝒚 𝒅𝒙𝟐 − 𝟑 𝒅𝒚 𝒅𝒙 + 𝟐𝒚 = 𝟎 −− −(𝟏) Let, y = 𝑒 𝑚𝑥 be the trail solution. ∴ 𝑑𝑦 𝑑𝑥 = 𝑚𝑒 𝑚𝑥 (diff.w.r.to x) ∴ 𝑑2 𝑦 𝑑𝑥2 = 𝑚2 𝑒 𝑚𝑥 (again diff.w.r.to x) From equation (1)- 𝑚2 𝑒 𝑚𝑥 - 3𝑚𝑒 𝑚𝑥 + 2𝑚𝑒 𝑚𝑥 =0 ⇒ 𝑒 𝑚𝑥 (𝑚2 − 3𝑚 + 2) = 0 ⇒ 𝑚2 − 3𝑚 + 2 = 0 ⇒ 𝑚2 − 2𝑚 − 𝑚 + 2 = 0 ⇒ m (m -2) -1 (m -2) = 0 ⇒ (m -2) (m-1) = 0 ∴ m = 1, 2 Solution 6
  • 7.
    Roots are realand unequal. So the general solution is- y = 𝑐1 𝑒 𝑚1 𝑥 + 𝑐2 𝑒 𝑚2 𝑥 ∴ y = 𝑐1 𝑒 𝑥 + 𝑐2 𝑒2𝑥 −− −(2) ⇒ 2 = 𝑐1 𝑒0 + 𝑐2 𝑒0 (∵ 𝑥 = 0) ⇒ 2 = 𝑐1 + 𝑐2 −− − 3 Now from equation (2) – 𝑑𝑦 𝑑𝑥 = 𝑐1 𝑒 𝑥 + 2𝑐2 𝑒2𝑥 ⇒ 0 = 𝑐1 𝑒0 + 2𝑐2 𝑒0 ⇒ 0 = 𝑐1 + 2𝑐2 − − − (4) Now (3) – (4) 2 = 𝑐1 + 𝑐2 0 = 𝑐1 + 2𝑐2 2 = − 𝑐2 ⇒ 𝑐2 = −2 From equation (3) 2 = 𝑐1 + −2 ⇒ − 𝑐1= − 2 − 2 ⇒ 𝑐1= 4 Hence the general solution is – y = 4𝑒 𝑥 - 2𝑒2𝑥 (Ans:) 7
  • 8.