(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
$\displaystyle y(t)$ and $\displaystyle x=\sin(t)$
Then by the chain rule we have
$\displaystyle \frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}$
We can take the derivative again using the product and chain rule to get
$\displaystyle \frac{d^2y}{dx^2} =\left( \frac{dx}{dt}\right)^2\frac{d^2y}{dx^2}+\frac{dy}{ dx}\frac{d^2x}{dt^2}$
If you work the above derivatives you will get the first two terms of your ODE.
Best wishes.
TES
My question is
$\displaystyle (1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} + 4y = 2x \sqrt{1-x^2} $
The sub is
$\displaystyle x = \sin t \Rightarrow \dfrac{dx}{dt} = \cos t \Rightarrow \dfrac{d^2x}{dt^2} = -\sin t $
$\displaystyle \dfrac{dy}{dt} = \dfrac{dy}{dx} \dfrac{dx}{dt} $
I made a mistake in follwing, i write it like below but when i differentiate with respect to t,
$\displaystyle \dfrac{dy}{dx} = \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}} $
the left hand side I write it like this
$\displaystyle \dfrac{d^2y}{dx^2} $ which should be like this $\displaystyle \dfrac{d^2y}{dx^2} \dfrac{dy}{dt} $
Thanks very much both