I am solving an ODE
$$y^2 \, dx + \left(x\sqrt{y^2 – x^2} – xy\right)dy=0$$
I let $y=xu$ and did some algebra, and I ended up with this
$$\ln(x)+C=\int \frac{1}{\sqrt{u^2 – 1}} \, du – \int \frac{1}{u} \, du$$
I don't know how to solve the first right hand side integral. I tried to let $u^2 – 1 = t$, so it becomes
$$\int\frac{1}{\sqrt{u^2 – 1}}=\int\frac{1}{\sqrt{t}}\frac{1}{2u} \, dt = \int \frac{1}{\sqrt{t}} \frac{1}{2} \frac{1}{\sqrt{t-1}} \, dt,$$
which doesn't seem to work. The correct answer seems to be
$$\ln \left(\sqrt{x^2 – 1} – x\right)$$
But how do we get this?
Best Answer
$$\int \frac { dx }{ \sqrt { x^{ 2 }-1 } } =\int { \frac { \sinh { t } }{ \sinh { t } } dt=t } +C\\ \\ x=\cosh { t } \\ dx=\sinh { t } \\ x=\frac { { e }^{ t }+{ e }^{ -t } }{ 2 } \\ 2x{ e }^{ t }={ e }^{ 2t }+1\\ { e }^{ 2t }-2x{ e }^{ t }+1=0\\ { e }^{ t }=x+\sqrt { { x }^{ 2 }-1 } $$
$$t=\ln { \left| x+\sqrt { { x }^{ 2 }-1 } \right| } $$