(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
(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
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