Second linear differential euqation with variable coefficient how to solve

(1-x^2)y'' - xy' +4y = 2x (1 -x^2)^1/2

Hint use the substitution x =sin t

I used it and end with

cos t y'' + sin t y' - (sin t)/(cos t) y' + 4y = 2 sin t (cos t)

how to solve this i just want the name of the method

Re: Second linear differential euqation with variable coefficient how to solve

Hi !

There is a mistake in your substitution.

Show the details of what you have done and it will be possible to see where is the snag.

Re: Second linear differential euqation with variable coefficient how to solve

Re: Second linear differential euqation with variable coefficient how to solve

Quote:

Originally Posted by

**TheEmptySet** If looks like you just made an error in your dervaitves. You are really close!

Suppose that y is a function of t

and

Then by the chain rule we have

We can take the derivative again using the product and chain rule to get

If you work the above derivatives you will get the first two terms of your ODE.

Best wishes.

TES

My question is

The sub is

I made a mistake in follwing, i write it like below but when i differentiate with respect to t,

the left hand side I write it like this

which should be like this

Thanks very much both

Re: Second linear differential euqation with variable coefficient how to solve

Quote:

Originally Posted by

**Amer** which should be like this

No, that is your mistake.

In order to obtain d²y/dx² you have to write :

d²y/dx² = dz/dx = (dz/dt)/(dx/dt) where z = dy/dx

Frist, express dy/dx as an expression which contains only (dy/dt) and t , without any x in it.