# University of Florida/Egm4313/s12.team11.perez.gp/R3.2

#### Problem Statement

Developing the second homogeneous solution for the case of double real root as a limiting case of distinct roots.

#### Given

Consider two distinct roots of the form:

${\displaystyle \lambda _{1}=x\!}$ and ${\displaystyle \lambda _{2}=x+\epsilon \!}$

(where ${\displaystyle \epsilon \!}$ is perturbation).

#### Given

Find the homogeneous L2-ODE-CC having the above distinct roots.

#### Solution

${\displaystyle (\lambda -\lambda _{1})(\lambda -(\lambda _{1}))=0\!}$

${\displaystyle (\lambda -x)(\lambda -(x+\epsilon ))=0\!}$

${\displaystyle \lambda ^{2}-\lambda x-\lambda \epsilon -\lambda x+x^{2}+x\epsilon =0\!}$

${\displaystyle \lambda ^{2}-\lambda (2x+\epsilon )+x(x+\epsilon )=0\!}$

${\displaystyle \therefore y''-y'(2\lambda +\epsilon )+y\lambda (\lambda +\epsilon )=0\!}$ (1)

#### Given

Show that ${\displaystyle {\frac {e^{(\lambda +\epsilon )x}-e^{\lambda x}}{\epsilon }}\!}$ is a homogeneous solution. (2)

#### Solution

Let's find the corresponding derivatives:

${\displaystyle y={\frac {e^{(\lambda +\epsilon )x}-e^{\lambda x}}{\epsilon }}\!}$

${\displaystyle y'={\frac {(\lambda +\epsilon )e^{\lambda +\epsilon x}-\lambda e^{\lambda x}}{\epsilon }}\!}$

${\displaystyle y''={\frac {(\lambda +\epsilon )^{2}e^{(\lambda +\epsilon )x}-\lambda ^{2}e^{\lambda x}}{\epsilon }}\!}$

If we now take these three equations and plug them into the homogeneous L2-ODE-CC (1), we get:

${\displaystyle e^{(\lambda +\epsilon )x}(\lambda ^{2}+2\lambda \epsilon +\epsilon ^{2}-2\lambda ^{2}-2\lambda \epsilon -\lambda \epsilon -\epsilon ^{2}+\lambda ^{2}+\epsilon \lambda )+e^{\lambda x}(-\lambda ^{2}+2\lambda ^{2}+\lambda \epsilon -\lambda ^{2}-\lambda \epsilon )=0\!}$

${\displaystyle \therefore 0\equiv 0.\!}$

Since the left and right hand sides of the equation are zero, the solution is in fact a homogeneous equation.

#### Given

Find the limit of the homogeneous solution in (2) as epsilon approaches zero (think l'Hopital's Rule).

#### Solution

Using l'Hopital's Rule,

${\displaystyle \lim _{\epsilon \rightarrow 0}{\frac {e^{(\lambda +\epsilon )x}-e^{\lambda x}}{\epsilon }}={\frac {0}{0}}\!}$

(this is an indeterminate form).

L'Hopital's Rule states that we can divide this function into two functions, ${\displaystyle f(\epsilon )\!}$ and ${\displaystyle g(\epsilon )\!}$, and then find their derivatives and attempt to find the limit of ${\displaystyle {\frac {f'(\epsilon )}{g'(\epsilon )}}\!}$. If a limit exists for this, then a limit exists for our original function.

${\displaystyle \lim _{\epsilon \rightarrow 0}{\frac {f'(\epsilon )}{g'(\epsilon )}}=\lim _{\epsilon \rightarrow 0}{\frac {xe^{x(\lambda +\epsilon )}}{1}}\!}$

${\displaystyle ={\frac {xe^{(\lambda +0)x}}{1}}\!}$

${\displaystyle =xe^{x\lambda }.\!}$

#### Given

Take the derivative of ${\displaystyle e^{\lambda x}}$ with respect to lambda.

#### Solution

Taking the derivative with respect to lambda, we find that:

${\displaystyle {\frac {d(e^{\lambda x})}{d\lambda }}=xe^{\lambda x}}$.

It is important to remember that we must hold ${\displaystyle x\!}$ as a constant when finding this derivative.

#### Given

Compare the results in parts (3) and (4), and relate to the result by using variation of parameters

#### Solution

Taking a closer look at Parts 3 and 4 of this problem, we discover that they're in fact equal:

${\displaystyle {\frac {d(e^{\lambda x})}{d\lambda }}=\lim _{e\rightarrow 0}{\frac {e^{(\lambda +\epsilon )x}-e^{\lambda x}}{\epsilon }}=xe^{\lambda x}\!}$

#### Given

Numerical experiment: Compute (2) setting lambda equal to 5 and epsilon equal to 0.001 [/itex], and compare to the value obtained from the exact second homogeneous solution.

#### Solution

After performing these calculations, from (2) we get 148.478.

And from the exact second homogeneous solution, we get 200.05.