# LSO – Two equal real roots

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So we have one root, q. Let us say that we have two roots and the second 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 least as  ε2x2. We can get that from the series expansion of ex or 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 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 eqx,

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

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 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 too, and the suggested solution is a solution indeed.

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Previous page : LSO -Two different real roots
Next page : LSO - Two complex rootsLast modified: Dec 28, 2023 @ 17:36