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Overview An Example Double Check Discussion
Reduction of Order
Bernd Schr¨
oder
Bernd Schr¨
oder Louisiana TechUniversity, College of Engineering and Science
Reduction of Order
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Reduction of Order: Finding Second Solutions for Differential Equations, Study Guides, Projects, Research of Differential Equations
A step-by-step guide on how to use the Reduction of Order method to find a second solution for the given differential equation: x2y′′ −2y = 0, with y1 = x2. the calculations and discussions on the process.
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2021/2022
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Reduction of Order
Bernd Schr¨oder
Bernd Schr¨oder Louisiana Tech University, College of Engineering and Science
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What is Reduction of Order?
Bernd Schr¨oder Louisiana Tech University, College of Engineering and Science
logo
What is Reduction of Order?
- Typically, reduction of order is applied to second order linear differential equations of the form y′′^ + P(x)y′^ + Q(x)y = 0.
- We must already have one solution y 1 of the equation.
Bernd Schr¨oder Louisiana Tech University, College of Engineering and Science
logo
What is Reduction of Order?
- Typically, reduction of order is applied to second order linear differential equations of the form y′′^ + P(x)y′^ + Q(x)y = 0.
- We must already have one solution y 1 of the equation.
- Reduction of order assumes there is a second, linearly independent solution of a the form y = uy 1.
Bernd Schr¨oder Louisiana Tech University, College of Engineering and Science
logo
What is Reduction of Order?
- Typically, reduction of order is applied to second order linear differential equations of the form y′′^ + P(x)y′^ + Q(x)y = 0.
- We must already have one solution y 1 of the equation.
- Reduction of order assumes there is a second, linearly independent solution of a the form y = uy 1.
- There are two ways to proceed. 4.1 We can substitute y = uy 1 into the equation. This leads to a first order differential equation, which explains the name.
Bernd Schr¨oder Louisiana Tech University, College of Engineering and Science
logo
What is Reduction of Order?
- Typically, reduction of order is applied to second order linear differential equations of the form y′′^ + P(x)y′^ + Q(x)y = 0.
- We must already have one solution y 1 of the equation.
- Reduction of order assumes there is a second, linearly independent solution of a the form y = uy 1.
- There are two ways to proceed. 4.1 We can substitute y = uy 1 into the equation. This leads to a first order differential equation, which explains the name. 4.2 Or, we can use the formula y 2 = y 1
∫ (^) e− ∫^ P dx y^21 dx,^ which is obtained by doing the above substitution symbolically.
Bernd Schr¨oder Louisiana Tech University, College of Engineering and Science
logo
Use Reduction of Order to Find a Second
Solution for x^2 y′′^ − 2 y = 0, given that y 1 = x^2.
y = ux^2
Bernd Schr¨oder Louisiana Tech University, College of Engineering and Science
logo
Use Reduction of Order to Find a Second
Solution for x^2 y′′^ − 2 y = 0, given that y 1 = x^2.
y = ux^2 y′^ = u′x^2 + u 2 x
Bernd Schr¨oder Louisiana Tech University, College of Engineering and Science
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Use Reduction of Order to Find a Second
Solution for x^2 y′′^ − 2 y = 0, given that y 1 = x^2.
y = ux^2 y′^ = u′x^2 + u 2 x y′′^ = u′′x^2 + u′ 4 x + u 2 x^2
u′′x^2 + u′ 4 x + u 2
− 2 ux^2 = 0
Bernd Schr¨oder Louisiana Tech University, College of Engineering and Science
logo
Use Reduction of Order to Find a Second
Solution for x^2 y′′^ − 2 y = 0, given that y 1 = x^2.
y = ux^2 y′^ = u′x^2 + u 2 x y′′^ = u′′x^2 + u′ 4 x + u 2 x^2
u′′x^2 + u′ 4 x + u 2
− 2 ux^2 = 0 u′′x^4 + u′ 4 x^3 + u 2 x^2 − 2 ux^2 = 0
Bernd Schr¨oder Louisiana Tech University, College of Engineering and Science
logo
Use Reduction of Order to Find a Second
Solution for x^2 y′′^ − 2 y = 0, given that y 1 = x^2.
y = ux^2 y′^ = u′x^2 + u 2 x y′′^ = u′′x^2 + u′ 4 x + u 2 x^2
u′′x^2 + u′ 4 x + u 2
− 2 ux^2 = 0 u′′x^4 + u′ 4 x^3 + u 2 x^2 − 2 ux^2 = 0 u′′x^4 + u′ 4 x^3 = 0 u′′x + u′ 4 = 0
Bernd Schr¨oder Louisiana Tech University, College of Engineering and Science
logo
Use Reduction of Order to Find a Second
Solution for x^2 y′′^ − 2 y = 0, given that y 1 = x^2.
y = ux^2 y′^ = u′x^2 + u 2 x y′′^ = u′′x^2 + u′ 4 x + u 2 x^2
u′′x^2 + u′ 4 x + u 2
− 2 ux^2 = 0 u′′x^4 + u′ 4 x^3 + u 2 x^2 − 2 ux^2 = 0 u′′x^4 + u′ 4 x^3 = 0 u′′x + u′ 4 = 0 v := u′
Bernd Schr¨oder Louisiana Tech University, College of Engineering and Science
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Use Reduction of Order to Find a Second
Solution for x^2 y′′^ − 2 y = 0, given that y 1 = x^2.
dv dx =^ −
4 v x
Bernd Schr¨oder Louisiana Tech University, College of Engineering and Science
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Use Reduction of Order to Find a Second
Solution for x^2 y′′^ − 2 y = 0, given that y 1 = x^2.
dv dx =^ −
4 v x dv v =^ −^4
dx x
Bernd Schr¨oder Louisiana Tech University, College of Engineering and Science