Respuesta :

LammettHash
The solution depends on the value of [tex]k[/tex]. To make things simple, assume [tex]k>0[/tex]. The homogeneous part of the equation is

[tex]\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0[/tex]

and has characteristic equation

[tex]r^2-16k=0\implies r=\pm4\sqrt k[/tex]

which admits the characteristic solution [tex]y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}[/tex].

For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be [tex]y_p=ae^{4x}+be^x[/tex]. Then

[tex]\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x[/tex]

So you have

[tex]16ae^{4x}+be^x-16k(ae^{4x}+be^x)=9.6e^{4x}+30e^x[/tex]
[tex](16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x[/tex]

This means

[tex]16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}[/tex]
[tex]b(1-16k)=30\implies b=\dfrac{30}{1-16k}[/tex]

and so the general solution would be

[tex]y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x[/tex]

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