Respuesta :

caylus
Hello,

This exercice is very hard for me. I have found the theory in "Mathematics for students of engineering and applied science" by L.B.Benny (Oxford University Press) page 415"
[tex]x=e^t\\ \frac{dt}{dx}= \frac{1}{x}\\ x^2 \frac{d^2y}{dx^2}= \frac{d^2y}{dt^2} -\frac{dy}{dt} \\ [/tex]

[tex] \frac{d^2y}{dt^2} - \frac{dy}{dt} -7 \frac{dy}{dt} +16y=0\\ \frac{d^2y}{dt^2} -8 \frac{dy}{dt} +16y=0\\ \Delta=8^2-4*16=0\\ r=4\\ y=(a+bt)e^{4t} ==\ \textgreater \ y=(a+b\ ln(x))x^4 [/tex]

If i am wrong, forget all this.


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