Find the general solution of the linear ordinary differential equation

Question

$y'= 6(y - 2.5) \tanh 1.5x$

The general solution of the differential equation is to be found using integrating factor. It is a linear ODE of first order. How do I do it?

Answer 1

The ODE is

$$y'(x) = 6(y(x)-\frac{5}{2})\tanh(\frac{3}{2} x)\tag{1}$$

Letting

$$y(x) = \frac{5}{2} + z(x)$$

gives

$$z'(x) = 6 \;z(x) \tanh(\frac{3}{2} x)$$

or

$$\frac{z'(x)}{z(x)} = 6 \;\frac{\cosh(\frac{3}{2} x)}{\sinh(\frac{3}{2} x))}$$

This can be written as

$$\frac{d}{dx} \log(z(x) = 6\frac{2}{3} \frac{d}{dx}\log( \cosh(\frac{3}{2})) x$$

Integrating yields

$$\log(z(x)) = 4\log( \cosh(\frac{3}{2} x))) + c$$

or, exponentiating (setting $A = e^c$)

$$z(x) = A \cosh(\frac{3}{2} x))^4$$

Hence the general solution of (1) is

$$y(x) = \frac{5}{2} + A \cosh(\frac{3}{2} x))^4$$

where A is an arbitrary constant.

Answer 2

Try separating the equation

$$ \frac{y'}{y-2.5} = \frac{6\sinh(1.5 x)}{\cosh(1.5 x)} $$

Integrating gives

$$ \ln|y-2.5| = 4\ln\big(\cosh(1.5 x)\big) + C $$

$$ \implies y = 2.5 + A\big(\cosh(1.5 x)\big)^4 $$