# Thread: Second linear differential euqation with variable coefficient how to solve

1. ## 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

2. ## 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.

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

Originally Posted by Amer
(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

$y(t)$ and $x=\sin(t)$

Then by the chain rule we have

$\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}$

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

$\frac{d^2y}{dx^2} =\left( \frac{dx}{dt}\right)^2\frac{d^2y}{dx^2}+\frac{dy}{ dx}\frac{d^2x}{dt^2}$

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

Best wishes.

TES

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

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

$y(t)$ and $x=\sin(t)$

Then by the chain rule we have

$\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}$

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

$\frac{d^2y}{dx^2} =\left( \frac{dx}{dt}\right)^2\frac{d^2y}{dx^2}+\frac{dy}{ dx}\frac{d^2x}{dt^2}$

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

Best wishes.

TES
My question is
$(1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} + 4y = 2x \sqrt{1-x^2}$

The sub is
$x = \sin t \Rightarrow \dfrac{dx}{dt} = \cos t \Rightarrow \dfrac{d^2x}{dt^2} = -\sin t$

$\dfrac{dy}{dt} = \dfrac{dy}{dx} \dfrac{dx}{dt}$

I made a mistake in follwing, i write it like below but when i differentiate with respect to t,
$\dfrac{dy}{dx} = \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}$
the left hand side I write it like this
$\dfrac{d^2y}{dx^2}$ which should be like this $\dfrac{d^2y}{dx^2} \dfrac{dy}{dt}$

Thanks very much both

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

Originally Posted by Amer
$\dfrac{d^2y}{dx^2}$ which should be like this $\dfrac{d^2y}{dx^2} \dfrac{dy}{dt}$
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.