LSO - Two equal real roots -

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So we have one root, q. Let us say that we actually have two roots where the other one is close to the first. So p=q+ε is another root where ε is a small number. We have then

$\begin{gathered} y = C{e^{px}} + D{e^{qx}} = C{e^{(q + \varepsilon )x}} + D{e^{qx}} \hfill \\ \quad = C{e^{qx}}{e^{\varepsilon x}} + D{e^{qx}} = {e^{qx}}(C{e^{\varepsilon x}} + D) \hfill \\ \end{gathered}$

as our solution. We then have that

${e^{\varepsilon x}} = 1 + \varepsilon x + ...$

where the … represent things that are varying at lest as ε²x². We can get that from the series expansion of e^x or from a reasoning around linear approximations of functions. So for small values of ε we get that

$\begin{gathered} y = {e^{qx}}(C{e^{\varepsilon x}} + D) \hfill \\ \quad \approx {e^{qx}}(C(1 + \varepsilon x) + D) \hfill \\ \quad = {e^{qx}}(C + C\varepsilon x + D) \hfill \\ \end{gathered}$

Now let C+D be our new D and Cε be our new C. Remember that our old C can be chosen arbitrarily large as ε gets smaller and smaller, so the new C can in the end have any value. This gives us, as ε goes toward 0

$y = {e^{qx}}(Cx + D)$

as our general solution. The above is in itself not a proof, but we can make it a proof (at least that the above represents solutions) by substituting it into our original equation. We have that

$\begin{gathered} y' = q{e^{qx}}(Cx + D) + C{e^{qx}} \hfill \\ \quad = {e^{qx}}(Cqx + Dq + C) \hfill \\ \end{gathered}$

and

$\begin{gathered} y'' = q{e^{qx}}(Cqx + Dq + C) \hfill \\ \quad = {e^{qx}}(C{q^2}x + D{q^2} + 2Cq) \hfill \\ \end{gathered}$

This gives us

${e^{qx}}(C{q^2}x + D{q^2} + 2Cq) + a{e^{qx}}(Cqx + Dq + C) + b{e^{qx}}(Cx + D) = 0$

or, removing the common factor e^qx,

$C{q^2}x + D{q^2} + 2Cq + aCqx + aDq + aC + bCx + bD = 0$

This can be written as

$\begin{gathered} C{q^2}x + D{q^2} + 2Cq + aCqx + aDq + aC + bCx + bD = 0 \hfill \\ Cx({q^2} + aq + b) + D({q^2} + aq + b) + C(2q + a) = 0 \hfill \\ \end{gathered}$

The two first terms must obviously be 0 since q is indeed the root of the characteristic equation. Also, since it is a double root we have that

${r^2} + ar + b = 0 = {(r - q)^2} = {r^2} - 2qr + {q^2}$

So a=–2q and therefore is the last term zero to, and the suggested solution is a solution indeed.

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Last modified: Mar 26, 2019 @ 21:08