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R1.4

Report 1: Problem R1.4

R2.5
Contributed by Charles Chiamchittrong.

Problem Statement:
Kreyszig 2011 p.59 pbs. 16-17

We are asked to find a second-order homogeneous linear ODE of the form


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! $$ \ y'' + ay' + by= 0 $$ !!  (Eq. 1) for a given basis of solutions.
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Problem 16
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! $$ \ e^{2.6x} \, \ e^{-4.3x} $$ !! 
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Problem 17
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! $$ \ e^{-\sqrt{5}x} \, \ xe^{-\sqrt{5}x} $$ !! 
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Solution:
Problem 16

Examining the given basis, we can see that this is a case of two distinct real roots.

The general solution in this case is:
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! $$ \ y = c_1e^{\lambda_1x} + c_2e^{\lambda_2x} $$ !!  (Eq. 2)
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Substituting the basis of solutions, we obtain:
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! $$ \ y = c_1e^{2.6x} + c_2e^{-4.3x} $$ !!  (Eq. 3)
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Comparing Eq. 2 and Eq. 3 we can determine the roots $$\lambda_1$$ and $$\lambda_2$$:


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! $$ \ \lambda_1 = 2.6 \, \ \lambda_2 = -4.3 $$ !!  (Eq. 4a, 4b)
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Now we can write the characteristic equation:


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! $$ \ (\lambda - \lambda_1)(\lambda - \lambda_2) = 0 $$ !!  (Eq. 5)
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! $$ \ (\lambda - 2.6)(\lambda + 4.3) = 0 $$ !!  (Eq. 6)
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! $$ \ \lambda^2 + 1.7\lambda - 11.18 = 0 $$ !!  (Eq. 7)
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Finally, we can convert the characteristic equation into the corresponding homogeneous ODE:


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! $$ \ y'' + 1.7y' - 11.18y = 0 $$  (Eq. 8)
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Problem 17

Examining the given basis, we can see that this is a case of a real double root.

The general solution in this case is:
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! $$ \ y = (c_1 + c_2x)e^{\lambda x} $$ !!  (Eq. 9)
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Substituting the basis of solutions, we obtain:
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! $$ \ y = (c_1 + c_2x)e^{-\sqrt{5} x} $$ !!  (Eq. 10)
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Comparing Eq. 8 and Eq. 9 we can determine the root $$\lambda_1$$:


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! $$ \ \lambda_1 = -\sqrt{5} $$ !!  (Eq. 11)
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Now we can write the characteristic equation:


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! $$ \ (\lambda - \lambda_1)^2 = 0 $$ !!  (Eq. 12)
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! $$ \ (\lambda + \sqrt{5})^2 = 0 $$ !!  (Eq. 13)
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! $$ \ \lambda^2 + 2\sqrt{5}\lambda + 5 = 0 $$ !!  (Eq. 14)
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Finally, we can convert the characteristic equation into the corresponding homogeneous ODE:


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! $$ \ y'' + 2\sqrt{5}y' + 5y = 0 $$ <p style="text-align: right;"> (Eq. 15)
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