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 ex 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 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 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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Previous page : LSO -Two different real roots
Next page : LSO - Two complex rootsLast modified: Mar 26, 2019 @ 21:08