How to analytically solve a Second Order ODE? I got the following ODE in my work. I haven't seen ODEs carrying trigonometric functions this way. See if anyone could help on solving it analytically. Here is the ODE
$y''(x)+\sqrt{2} c \tanh \left(\dfrac{c x}{3 \sqrt{2}}\right) y'(x)=0$ where $c>0.$ The boundary conditions could be $y(\infty)=0, y(0)=1.$
 A: Letting $u(x)=y'(x)$ and $m={{c}\over 3\sqrt2}\space$yields;
$$u'(x)+{\sqrt{2}c \tanh(mx)}\cdot u(x)=0$$
Creating an integrating factor, $\mu(x)$, gives us;
$$\mu(x)=e^{\sqrt{2}c\int{\tanh(mx)}dx}=e^{6\cdot {\ln(\cosh(mx))}}$$
$$\therefore \mu(x)=(\cosh(mx))^{6}$$
Multiplying our equation by this integrating factor yields;
$$(\cosh(mx))^{6}\cdot u'+{\sqrt{2}c\tanh(mx)}\cdot(\cosh(mx))^{6}\cdot u=0$$
Which can be rewritten thusly;
$${d\over dx}\big((\cosh(mx))^{6}\cdot u(x)\big)=0$$
Integrating on both sides and a quick back-substitution reveals;
$$u(x)=y'(x)={C\over(\cosh(mx))^{6}}$$
$$\therefore y(x)=C\int{(\cosh(mx))^{{-6}}}dx$$
Here I must admit to a certain amount of laziness. This integral is quite doable but tedious in the extreme... The final result is;
$$y(x)=C \cdot {{\sqrt{2}\tanh(mx)\cdot(3\tanh^4(mx)-10\tanh^2(mx)+15)}\over5c}+D$$
Finally, applying your boundary conditions reveals that $(D=1)$ and $(C={-5c\over 8\sqrt{2}})$;
$$\therefore y(x)= {{-\tanh\big({{cx}\over 3\sqrt2}\big)\big(3\tanh^4\big({{cx}\over 3\sqrt2}\big)-10\tanh^2\big({{cx}\over 3\sqrt2}\big)+15\big)}\over  8}+1$$
