3

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?

Community
  • 1
3x89g2
  • 7,542

5 Answers5

10

$$\int \frac { dx }{ \sqrt { x^{ 2 }-1 } } =\int { \frac { \sinh { t } }{ \sinh { t } } dt=t } +C\\ \\ x=\cosh { t } \\ dx=\sinh { t } \\ x=\frac { { e }^{ t }+{ e }^{ -t } }{ 2 } \\ 2x{ e }^{ t }={ e }^{ 2t }+1\\ { e }^{ 2t }-2x{ e }^{ t }+1=0\\ { e }^{ t }=x+\sqrt { { x }^{ 2 }-1 } $$

$$t=\ln { \left| x+\sqrt { { x }^{ 2 }-1 } \right| } $$

haqnatural
  • 21,578
2

Let $x=\sec t, \; dx=\sec t\tan t\,dt$ to get

$\displaystyle\int\frac{1}{\sqrt{x^2-1}}dx=\int\frac{\sec t\tan t}{\tan t}dt=\int\sec t\, dt=\ln\big|\sec t+\tan t\big|+C=\ln\big|x+\sqrt{x^2-1}\big|+C$

user84413
  • 27,211
1

LEt $u=\cosh t$. You may need to find a formula for arccosh x

Empy2
  • 50,853
0

If the hyperbolic trigonometry is not known, we can see that

$\int\frac{dt}{\sqrt{t}\sqrt{t+1}}=\int\frac{\sqrt{t+1}}{\sqrt{t}}-\frac{\sqrt t}{\sqrt{t+1}}dt.$

Then $z=1+\frac{1}{t}$ gives $t=\frac{1}{z-1}$, $dt=\frac{-dz}{(1-z)^2}$ and

$\frac{1}{2}\int\frac{\sqrt{t+1}}{\sqrt{t}}-\frac{\sqrt t}{\sqrt{t+1}}dt=\frac{1}{2}\int\frac{(1-z)dz}{\sqrt{z}(1-z)^2}=\frac{1}{2}\int\frac{dz}{\sqrt{z}(1-z)}=\frac{1}{4}\int\frac{1}{\sqrt z(1-\sqrt{z})}+\frac{1}{\sqrt z(1+\sqrt{z})}dz.$

It comes

$\int\frac{du}{\sqrt{u^2-1}}=\frac{1}{2}\left(\ln(1+\sqrt z)-\ln(\sqrt z-1)\right)=\frac{1}{2}\ln\left(\frac{1+\sqrt z}{\sqrt z-1}\right)=\frac{1}{2}\ln\left(\frac{\sqrt t+\sqrt{1+t}}{\sqrt{1+t}-\sqrt t}\right)=\frac{1}{2}\ln\left(\frac{u+\sqrt{u^2-1}}{u-\sqrt{u^2-1}}\right)$

and as $\ln\left(\frac{1}{u-\sqrt{u^2-1}}\right)=\ln(u+\sqrt{u^2-1})$, we obtain $\int\frac{du}{\sqrt{u^2-1}}=\ln(u+\sqrt{u^2-1})$.

We can also note that $\int\frac{du}{\sqrt{u^2-1}}=\frac{1}{2}\int\frac{\sqrt{u+1}}{\sqrt{u-1}}-\frac{\sqrt{u-1}}{\sqrt{1+u}}du$ and take the substitution $t=\frac{u+1}{u-1}$.

calacala
  • 61
-2

Use a substitution of $u=\sin(t)$, differentiate to get $du$ then sub into that 2nd integral.

Then evaluate that integral & should progress towards the answer.

unseen_rider
  • 778