Re: 2nd order linear DE

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Posted by Subhotosh Khan on September 27, 2002 at 08:32:08:

In Reply to: 2nd order linear DE posted by colin on September 26, 2002 at 16:52:14:

: This is one of those 2nd order linear DEs where the independent variable is
: missing.
: After making a transformation using v = y' the eq is supposed to be a
: Bernoulli eq.

: y'' + (y')^2 = 2e^-y

: v' + v^2 = 2e^-y

: v' = dv/dt = (dv/dy)(dy/dt) = v(dv/dy)

: v(dv/dy) + v^2 = 2e^-y

: dv/dy + v = 2(v^-1)(e^-y)

: Is this a Bernoulli eq. with n = -1 ?

: If so, then I'm letting r = v^(1-n) = r = v^2
: dr/dy = 2v(dv/dy)

: v(dv/dy) + v^2 = 2e^-y
: 1/2(dr/dy) + r = 2e^-y
: r' + 2r = 4e^-y

: integrating factor = e^2y
: re^2y = 4e^y + c1
: r = 4e^-y +c1*e^-2y
: v = 4e^-y/2 + c1*e^-y
*******************************

r = 4e^-y + c1*e^-2y

v= r^(1/2)= (4e^-y + c1*e^-2y)^(1/2)

Integrating this could be difficult - probably you have to make another substitution like -y = ln x.

then

v = (4e^-y + c1*e^-2y)^(1/2)

v = (4x + c1*x^2)^(1/2)

v = dy/dt = dy/dx * dx/dt = (-1/x)*dx/dt

dx/[x * (4x + c1*x^2)^(1/2)] = -dt

Well.... this is messy
*********************************

: 4e^y/2 + c1*e^y dy = dt
: t + c2 = 8e^y/2 + c1*e^y

: can some help point out my screw up?? thx...

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