find the general solution of the following Bernoulli equation, dx⁄dt =αx^2/3−βx α, β ≠0

Question

find the general solution of the following Bernoulli equation, dx⁄dt =αx^2/3−βx

α, β ≠0

Answer 1

Apply the substitution y = x^1-2 = 1/x. Then dy/dt = -1/x² dx/dt. Multiply the equation by -1/x² (thus forcing dy/dt to show up), and get

-1/x² dx/dt = -α/3 + β/x, which in terms of the new variable y can be written as

dy/dt = -α/3 + βy, or equivalently dy/dt -βy = -α/3 (*)

This equation is linear, so an integrating factor is I(t) = e^∫-βdt = e^-βt. Multiplying across we get

d/dt(y e^-βt ) = -α e^-βt/3. Integrating we obtain

y e^-βt = (α e^-βt )/(3β) + C, so the general solution of (*) is

y = α/(3β) + Ce^βt.

Finally, substituting back, since x = 1/y we obtain

x(t) = 1/(α/(3β) + Ce^βt)