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.
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 $\displaystyle u = \frac{dy}{dx}$
$\displaystyle \frac{dy}{dx} = \frac{C}{x}$
$\displaystyle y = \int{\frac{C}{x}\,dx}$
$\displaystyle y = C\ln{|x|} + D$.
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
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