# Thread: [SOLVED] How do you solve y'' + (y')/x = 0

1. ## [SOLVED] How do you solve y'' + (y')/x = 0

I have the following homogenous 2nd order diff. eqn that has me stumped.

y'' + y'/x = 0

I don't know how to get the general soln. Any help would be greatly appreciated.

2. Originally Posted by spearfish I have the following homogenous 2nd order diff. eqn that has me stumped.

y'' + y'/x = 0

I don't know how to get the general soln. Any help would be greatly appreciated.
Let $\displaystyle y' = u$. The ODE will separate. Then integrate again.

3. Thanks Danny,

Ok, so if I let y' = u, then y'' = u', so I get u' + u/x = 0. Then I do u' = -u/x, Integrate to get U = -U^2/(2x).

Where do I go from here? I don't even know if what I did is valid? I am just really confused on this one.

4. Originally Posted by spearfish Thanks Danny,

Ok, so if I let y' = u, then y'' = u', so I get u' + u/x = 0. Then I do u' = -u/x, Integrate to get U = -U^2/(2x).

Where do I go from here? I don't even know if what I did is valid? I am just really confused on this one.
No. Separate $\displaystyle \frac{du}{u} = - \frac{dx}{x}$.

5. ahhh, now it's coming back to me. I ll work on in it some more and post back later. Thanks

6. Alternatively, you can use the Integrating Factor.

$\displaystyle \frac{du}{dx} + \frac{1}{x}u = 0$

Integrating Factor: $\displaystyle e^{\int{\frac{1}{x}\,dx}} = e^{\ln{x}} = x$.

Multiply both sides by the integrating factor:

$\displaystyle x\frac{du}{dx} + u = 0$

$\displaystyle \frac{d}{dx}(ux) = 0$

$\displaystyle ux = \int{0\,dx}$

$\displaystyle ux = C$

$\displaystyle u = \frac{C}{x}$.

Now since $\displaystyle u = \frac{dy}{dx}$

$\displaystyle \frac{dy}{dx} = \frac{C}{x}$

$\displaystyle y = \int{\frac{C}{x}\,dx}$

$\displaystyle y = C\ln{|x|} + D$.

7. As alternative You can use the ever useful change on variable $\displaystyle u=\ln x$ illustrated in...

http://www.mathhelpforum.com/math-he...-question.html

We have...

$\displaystyle \frac{dy}{dx} = \frac{1}{x}\cdot \frac{dy}{du}$

$\displaystyle \frac{d^{2}y}{dx^{2}}=\frac{1}{x^{2}}\cdot (\frac{d^{2}y}{du^{2}} - \frac{dy}{du})$ (1)

... so that ther DE...

$\displaystyle \frac{d^{2}y}{dx^{2}} + \frac{1}{x}\cdot \frac{dy}{dx}=0$ (2)

...becomes...

$\displaystyle \frac{d^{2}y}{du^{2}} =0$ (3)

The integration of (3) is immediate and we obtain...

$\displaystyle y=c_{1}\cdot u + c_{2} = c_{1}\cdot \ln x + c_{2}$ (4) Merry Christmas from Italy

$\displaystyle \chi$ $\displaystyle \sigma$

8. Thanks guys.

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