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

Quote:

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

$\displaystyle y(t)$ 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

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

$\displaystyle y(t)$ 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

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

Quote:

Originally Posted by

**Amer** $\displaystyle \dfrac{d^2y}{dx^2} $ which should be like this $\displaystyle \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.