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 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 y" + p(x)y +q(x)y=0_p(x) = a1(x) a₂(x)' 9(x) = ao(x) a₂(x)' Then the method of reduction of order gives a second linearly independent solution as y2(x) = Cy₁u = Cy₁ (x) e-/p(x)dx y₁(x) -dx 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 = C3e2x then we can choose C = 1/3 so that y2 = e2x. Given the problem y" - 4y' + 13y = 0 and a solution y₁ = e2x sin(3x) Applying the reduction of order method we obtain the following y₁(x) = p(x) = e and e-/p(x)dx = So we have e-/p(x)dx y₁√(x) dx = dx = Finally, after making a selection of a value for C as described above (you have to choose some nonzero numerical value) we arrive at So the general solution to y" -3y+ 4y = 0 can be written as y2(x) = Cy₁u = y= ciyi+C2y2 = c1 +c2
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 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 y" + p(x)y +q(x)y=0_p(x) = a1(x) a₂(x)' 9(x) = ao(x) a₂(x)' Then the method of reduction of order gives a second linearly independent solution as y2(x) = Cy₁u = Cy₁ (x) e-/p(x)dx y₁(x) -dx 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 = C3e2x then we can choose C = 1/3 so that y2 = e2x. Given the problem y" - 4y' + 13y = 0 and a solution y₁ = e2x sin(3x) Applying the reduction of order method we obtain the following y₁(x) = p(x) = e and e-/p(x)dx = So we have e-/p(x)dx y₁√(x) dx = dx = Finally, after making a selection of a value for C as described above (you have to choose some nonzero numerical value) we arrive at So the general solution to y" -3y+ 4y = 0 can be written as y2(x) = Cy₁u = y= ciyi+C2y2 = c1 +c2
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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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 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
y" + p(x)y +q(x)y=0_p(x) =
a1(x)
a₂(x)'
9(x) =
ao(x)
a₂(x)'
Then the method of reduction of order gives a second linearly independent solution as
y2(x) = Cy₁u = Cy₁ (x)
e-/p(x)dx
y₁(x)
-dx
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 = C3e2x then we can choose C = 1/3 so that y2 = e2x.
Given the problem
y" - 4y' + 13y = 0
and a solution y₁ = e2x sin(3x)
Applying the reduction of order method we obtain the following
y₁(x) =
p(x) = e
and e-/p(x)dx =
So we have
e-/p(x)dx
y₁√(x)
dx =
dx =
Finally, after making a selection of a value for C as described above (you have to choose some nonzero numerical value) we arrive at
So the general solution to y" -3y+ 4y = 0 can be written as
y2(x) = Cy₁u =
y= ciyi+C2y2 = c1
+c2
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