Integral of y*tanh(x) + constant ?

Question

How do I find the general solution of

dy/dx = y*tanh(x) - 2

I know tanh(x) = [exp(x) - exp(-x)] /[exp(x) + exp(-x)] among other things.

I don't know what to do with the 2 and what form of tanh(x) to use. Any help would be appreciated.

Answer 1

Write this as

dy/dx  - tanh(x) y = -2

This is first order linear. There is always an integrating factor; here it's

e^( ∫ -tanh(x) dx) = e^(-log(cosh(x)) = e^(log(sech(x)) = sech(x)

Multiply both sides by this factor:

sech(x) dy/dx - sech(x) tanh(x) y = -2 sech(x)

d/dx (sech(x) y) = -2 sech(x)

Integrating,

sech(x) y = ∫ [-4/(e^x + e^(-x))] dx

= -4 ∫ [e^x/(e^(2x) + 1)] dx

= -4 ∫ e^x/{e^(x)}² + 1] dx

= -4 arctan(e^x) + C   [Use the substitution u = e^x for this last step]

y = -4 cosh(x) arctan(e^x) + C cosh(x)