What is the Substitution for Solving Second Order Differential Equations?

[scottie_000]

I am having real trouble with this second order differential
The substitution is given and i just can't seem to use it
What am i missing here?

$$x \frac{d^2 y} {dx^2} -2 \frac{dy} {dx} + x = 0, \frac{dy} {dx} = v$$

All help welcome

 

[d_leet]

You need to show some work. What problems are you having with that substitution, it shouldn't be very hard to use.

 

[scottie_000]

That's just the problem. I can't get much at all.
The only step I can think of is:

$$x \frac {dv}{dx} + 2v + x = 0$$

I don't see where that leads me.
Do i need to find x in terms of v?...

 

[Valhalla]

From where you are put it in standard form and then think about how you would solve that ODE.

 

[scottie_000]

I've honestly tried for ages to put it in standard form.
I've tried finding dy/dv, dx/dv, dv/dy etc...
None of that has worked at all...

 

[d_leet]

That's just the problem. I can't get much at all.
The only step I can think of is:

$$x \frac {dv}{dx} + 2v + x = 0$$

I don't see where that leads me.
Do i need to find x in terms of v?...

No you need to find v in terms of x. Solve that relatively simple ODE for the function v and then you have another even simpler equation that if you solve will give you y.

you have

xv' + 2v + x = 0

so

xv' + 2v = -x

Can you solve that equation?

 

[HallsofIvy]

In your first post, you said the equation was
$$x \frac{d^2 y} {dx^2} -2 \frac{dy} {dx} + x = 0, \frac{dy} {dx} = v$$
Now, after substituting v for dy/dx, you write
$$x \frac {dv}{dx} + 2v + x = 0$$

One of those has a typo. Assuming the first equation is what you want, the second should be
$$x \frac {dv}{dx} - 2v + x = 0$$
or, as d leet said, xv'- 2v= -x (correcting that sign error).
That's a linear first order differential equation and has a simple integrating factor.

Another method: write the equation as xv'= 2v- x and divide by x:
v'= 2(v/x)- 1. Now let u= v/x so that v= xu. v'= xu'+ u and the equation becomes xu'+ u= 2u- 1 or xu'= u-1, a simple separable equation.

 

[scottie_000]

Ahhh thank you...
I couldn't see what i was missing
It's all so simple