# Thread: Solve (x^2) ( d^2/dx^2) + 3x(dy/dx) + y = 1/x

1. ## Solve (x^2) ( d^2/dx^2) + 3x(dy/dx) + y = 1/x

(x^2) ( d^2/dx^2) + 3x(dy/dx) + y = 1/x

how can i find the general solution of differential equation?

2. It's a Cauchy-Euler equation. Use that procedure.

3. but right side of equation isn't equal to zero? can you solve it for me?

4. 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.

5. 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.

6. 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.

7. Originally Posted by Ackbeet
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.
I am actually curious how to get the PI. I can get the CF. Would a first guess of PI=Ax^n be a good guess to suit 1/x?

8. Originally Posted by bugatti79
I am actually curious how to get the PI. I can get the CF. Would a first guess of PI=Ax^n be a good guess to suit 1/x?
Why not try it and see what happens?

9. 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!

10. Originally Posted by TheEmptySet
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?

11. @bugatti79

What I have is

$\displaystyle \frac{d}{dx}\left(x^2y'+xy \right)=\frac{d}{dx}\ln(x)$

Now if you integrate both sides you get

$\displaystyle x^2y'+xy =\ln(x)+C$

This is now a first order ODE that can be solve via an integrating factor!

12. Originally Posted by TheEmptySet
@bugatti79

What I have is

$\displaystyle \frac{d}{dx}\left(x^2y'+xy \right)=\frac{d}{dx}\ln(x)$

Now if you integrate both sides you get

$\displaystyle x^2y'+xy =\ln(x)+C$

This is now a first order ODE that can be solve via an integrating factor!
Yes I solved it using the integrating factor method. I am curious where I went wrong when guessing the PI in post #10. Where are you Ackbeet?

13. Originally Posted by bugatti79
Where are you, Ackbeet?
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.

14. Originally Posted by Ackbeet
Not paying attention to the fact that you tried my suggestion, apparently.

The next step with Cauchy-Euler equations is the (ln(x))^2 / x solution, so try a constant times that.
Hmmmm..I got the first and second A times the above and put back into DE. I arrived at 0=1. Not to worry, I learned a nice bit and some differentiating practice. At least we know the integrating factor works. :-)

15. Originally Posted by bugatti79
Hmmmm..I got the first and second A times the above and put back into DE. I arrived at 0=1. Not to worry, I learned a nice bit and some differentiating practice. At least we know the integrating factor works. :-)
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

$\displaystyle y_{p}=\frac{\ln^{2}(x)}{2x}.$

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