Given a second order linear homogeneous differential equation a₂(x)y" + a₁(x)y' + a₁(x)y=0 we know that a fundamental set for this ODE consists of a pair linearly independent solutions y₁, y2. But there are times when only one function, call it y₁, is available and we would like to find a second linearly independent solution. We can find y2 using the method of reduction of order. First, under the necessary assumption the a₂(x) #0 we rewrite the equation as Then the method of reduction of order gives a second linearly independent solution as and a solution y₁ = 2¹ Applying the reduction of order method to this problem we obtain the following So we have p(x) = y" + p(x)y' +q(x)y=0 p(x) = Sp(z)dz [= So the general solution to 9y" - 3y + 4y = 0 can be written as where C' is an arbitrary constant. We can choose the arbitrary constant to be anything we like. One useful choice is to choose C so that all the constants in front reduce to 1. For example, if we obtain Y2 = C'3e² then we can choose C = 1/3 so that y₂ = ²¹. Given the problem - dx = Y2(x) = Cy₁u = Cy₁ (x) y₁(x) = a₁(x) a₂(x)' 2) [²√² (2) and e-Sp(z)dz x²y" + 3xy - 24y = 0 y = C₁Y₁+C232 C₁ x^4 Sp(z)dz q(x) = dx = -dx y} (x) Finally, after making a selection of a value for C as described above (you have to choose some nonzero numerical value) we arrive at y2(x) = Cy₁u = +0₂ a₁(x) a₂(x)'

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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Given a second order linear homogeneous differential equation
a₂(x)y" + a₁(x)y' + ao(x)y=0
we know that a fundamental set for this ODE consists of a pair linearly independent solutions 9₁, 92. But there are times when only one function, call it y₁, is available and we would like to find a second linearly
independent solution. We can find y2 using the method of reduction of order.
First, under the necessary assumption the a₂(x) #0 we rewrite the equation as
Then the method of reduction of order gives a second linearly independent solution as
and a solution ₁ = x¹
Applying the reduction of order method to this problem we obtain the following
So we have
y" + p(x)y' +q(x)y=0 p(x) =
p(x) =
[²
So the general solution to 9y" - 3y + 4y = 0 can be written as
-dx =
S
Y₂(x) = Cy₁u = Cy₁ (x)
where C' is an arbitrary constant. We can choose the arbitrary constant to be anything we like. One useful choice is to choose C' so that all the constants in front reduce to 1. For example, if we obtain
Y2 = C3e²z then we can choose C = 1/3 so that y₂ = e²1.
Given the problem
y(x) =
2) / ²
a₁(x)
a₂(x)'
and e-SP(z)dz
y = C₁y₁ + C₂Y2 = C₁ x^4
e-SP(x)dz
x²y" + 3xy' - 24y = 0
y²(x)
q(x) =
e-√p(z)dz
y} (x)
Finally, after making a selection of a value for C as described above (you have to choose some nonzero numerical value) we arrive at
y2(x) = Cy₁u =
dx =
dr
+0₂
a₁(x)
a₂(x)'
Transcribed Image Text:Given a second order linear homogeneous differential equation a₂(x)y" + a₁(x)y' + ao(x)y=0 we know that a fundamental set for this ODE consists of a pair linearly independent solutions 9₁, 92. But there are times when only one function, call it y₁, is available and we would like to find a second linearly independent solution. We can find y2 using the method of reduction of order. First, under the necessary assumption the a₂(x) #0 we rewrite the equation as Then the method of reduction of order gives a second linearly independent solution as and a solution ₁ = x¹ Applying the reduction of order method to this problem we obtain the following So we have y" + p(x)y' +q(x)y=0 p(x) = p(x) = [² So the general solution to 9y" - 3y + 4y = 0 can be written as -dx = S Y₂(x) = Cy₁u = Cy₁ (x) where C' is an arbitrary constant. We can choose the arbitrary constant to be anything we like. One useful choice is to choose C' so that all the constants in front reduce to 1. For example, if we obtain Y2 = C3e²z then we can choose C = 1/3 so that y₂ = e²1. Given the problem y(x) = 2) / ² a₁(x) a₂(x)' and e-SP(z)dz y = C₁y₁ + C₂Y2 = C₁ x^4 e-SP(x)dz x²y" + 3xy' - 24y = 0 y²(x) q(x) = e-√p(z)dz y} (x) Finally, after making a selection of a value for C as described above (you have to choose some nonzero numerical value) we arrive at y2(x) = Cy₁u = dx = dr +0₂ a₁(x) a₂(x)'
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