Second linear differential euqation with variable coefficient how to solve

**Amer** · October 9th 2012, 08:22 AM

(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

**JJacquelin** · October 9th 2012, 09:26 AM

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.

**TheEmptySet** · October 9th 2012, 11:07 AM

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

**Amer** · October 9th 2012, 09:41 PM

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

**JJacquelin** · October 9th 2012, 10:19 PM

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.

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