(x^2) ( d^2/dx^2) + 3x(dy/dx) + y = 1/x
how can i find the general solution of differential equation?
It's a Cauchy-Euler equation. Use that procedure.
We don't "solve things" for you. I would suggest either assuming a solution that looks like the RHS, or, if you choose the exponential approach (thus producing a linear DE with constant coefficients), the choice of a particular solution might be more obvious. Try those, and post your work, and we can go from there.
Just to add my 2 cents. If you can solve the homogenous equation (Ackbeet told you how) you can find the particular solution by using the variation of parameters.
Variation of parameters - Wikipedia, the free encyclopedia This method does not involve guessing but you may end up with integrals in your solution, but that wont be the case in this particular problem.
Put $\displaystyle t=\ln x$ so we have $\displaystyle \frac{dy}{dx}=\frac{dy}{dt}\cdot \frac{dt}{dx}=\frac{1}{x}\cdot \frac{dy}{dt},$ and then $\displaystyle \frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{d}{dx}\left( \frac{1}{x}\cdot \frac{dy}{dt} \right)=-\frac{1}{{{x}^{2}}}\cdot \frac{dy}{dt}+\frac{1}{x}\cdot\frac{{{d}^{2}}y}{d{ {t}^{2}}},$ so you'll end up with an ODE with constant coefficients.
I just noticed something that I think is worth mentioning.
Notice that on the left hand side we have
$\displaystyle x^2y''+3xy'+y=x^2y''+2xy'+xy'+y$
Notice that this is a perfect derivative
$\displaystyle \frac{d}{dx}\left(x^2y'+xy \right)$
So we can rewrite the equation as
$\displaystyle \frac{d}{dx}\left(x^2y'+xy \right)=\frac{d}{dx}\ln(x)$
and reduce this to a first order ODE!
Thats interesting. I see what you have done but how does one reduce it to a first order? I havent seen this before to the best of my poor memory :-) Anyhow, I attempted it doing it the hard way by guessing the PI. I dont think it works out....
$\displaystyle \displaystyle Let y=Ax^n, y'=nAx^{n-1}, y''=n(n-1)Ax^{n-2}$
Putting back into the original DE, I arrive at
$\displaystyle Ax^n[n^2+2n+1]=\frac{1}{x} \implies Ax^n=\frac{1}{x} \because [n^2+2n+1]=0 \therefore A=1$
Is this right?
Not paying attention to the fact that you tried my suggestion, apparently.
Yeah, so that particular solution doesn't work, does it? You've got 0 = 1/x, a contradiction. Apparently, we need to change the ansatz. Note that the complimentary solution actually contains 1/x in it already, and by variation of parameters, you can find out that the complimentary solution also contains ln(x) / x. The next step with Cauchy-Euler equations is the (ln(x))^2 / x solution, so try a constant times that.
Well, here's my shot at it.
$\displaystyle y=A\,\frac{\ln^{2}(x)}{x}$
$\displaystyle y'=A\,\frac{2(x)(\ln(x)/x)-(\ln^{2}(x))(1)}{x^{2}}=A\,\frac{2\ln(x)-\ln^{2}(x)}{x^{2}}.$
$\displaystyle y''=A\,\frac{(x^{2})(2/x-2\ln(x)/x)-2(2\ln(x)-\ln^{2}(x))(x)}{x^{4}}=A\,\frac{x(2-2\ln(x))-2x(2\ln(x)-\ln^{2}(x))}{x^{4}}$
$\displaystyle =2A\,\frac{1-3\ln(x)+\ln^{2}(x)}{x^{3}}.$
Plugging these into the DE yields
$\displaystyle 2A\,\frac{1-3\ln(x)+\ln^{2}(x)}{x}+3A\,\frac{2\ln(x)-\ln^{2}(x)}{x}+A\,\frac{\ln^{2}(x)}{x}=\frac{1}{x} \quad\implies$
$\displaystyle 2A=1,$ or $\displaystyle A=1/2.$
So the particular solution is
$\displaystyle y_{p}=\frac{\ln^{2}(x)}{2x}.$