Differentiate hyperbolic function

[DryRun]

Homework Statement
If lny = sinh^(-1)(x), prove that

(1+ x^2)y'' + xy' - y = 0

The attempt at a solution

I have tried various (unsuccessful) ways of doing this, but the basic procedure that I've done is:

D.w.r.t.x for lny = sinh^(-1)(x)

This gives: (1/y)y' = 1/(1 + x^2)

To obtain y'', d.w.r.t.x for equation above.

This is the part where I'm not sure if i should use product rule for differentiating (1/y)y'

Is is correct?
-y^(-2)y' + y^(-1)y''

Or is this the right way of doing it?
-y^(-2)y''

 

[LCKurtz]

This is the part where I'm not sure if i should use product rule for differentiating (1/y)y'

Is is correct?
-y^(-2)y' + y^(-1)y''

Or is this the right way of doing it?
-y^(-2)y''

You must use the product (or quotient) rule.

 

[Quinzio]

Well, you should be aware that:
sinh^{-1}x=log(x+\sqrt{x^2+1})

so you have
\log y=\log(x+\sqrt{x^2+1})

from now on it should be simple.

 

[vela]

You differentiate sinh^-1 x incorrectly. You should have
\frac{y'}{y} = \frac{1}{\sqrt{1+x^2}}Then I'd get rid of the quotients to make things simpler:
y'\sqrt{1+x^2} = ybefore differentiating. (I just never really liked the quotient rule.)

 

[DryRun]

I have taken into account all of your advices. OK, now here's the evolution of this problem from my side.

So, making y' the subject of formula, i get y' = y/[(1 + x^2)^(1/2)]

I further differentiate w.r.t.x the equation by vela above (i apologize for not using the better-looking LaTex, but i still have a lot to learn on how to use it):

So, y''(1 + x^2)^(1/2) + y'/[2(1 + x^2)^(1/2)] = y'

Now, i make y'' the subject of formula.

y'' = {y' - y'/[2(1 + x^2)^(1/2)]}/[(1 + x^2)^(1/2)]

Then, i replace y' and y'' above, into the original equation: (1+ x^2)y'' + xy' - y, hoping that it will all reduce to 0. After substituting, i will end up with a (rather long) equation involving x, y and y' (which i have to substitute again). Is this the right direction to solve this problem?

 

[vela]

I have taken into account all of your advices. OK, now here's the evolution of this problem from my side.

So, making y' the subject of formula, i get y' = y/[(1 + x^2)^(1/2)]

I further differentiate w.r.t.x the equation by vela above (i apologize for not using the better-looking LaTex, but i still have a lot to learn on how to use it):

So, y''(1 + x^2)^(1/2) + y'/[2(1 + x^2)^(1/2)] = y'

You need to use the chain rule when calculating the second term.

Now, i make y'' the subject of formula.

y'' = {y' - y'/[2(1 + x^2)^(1/2)]}/[(1 + x^2)^(1/2)]

Then, i replace y' and y'' above, into the original equation: (1+ x^2)y'' + xy' - y, hoping that it will all reduce to 0. After substituting, i will end up with a (rather long) equation involving x, y and y' (which i have to substitute again). Is this the right direction to solve this problem?

You're making more work for yourself than you need to.

Hint: After you differentiated, the y'' term almost looks like what you want.

 

[DryRun]

OK, i got it. Thank you all very much.