2nd order DE (non linear I think but homogeneous)

**arbolis** · August 1st 2010, 04:48 PM

Find the general solution of $x^2y''-5xy'+9y=0$ . Hint: Once known a solution $y_1(x)$ , another solution is given by $y(x)=v(x)y_1(x)$ , with $v(x)$ to determinate.
Attempt: I just don't know what method to use. I don't think variation of parameters work here since it's non linear.

**Chris L T521** · August 1st 2010, 05:11 PM

Its a Cauchy-Euler equation.

So suppose that the equation $ax^2y^{\prime\prime}+bxy^{\prime}+cy=0$ has a solution of the form $y=x^r$ . Plugging this back into the ODE yields the characteristic equation $ar^2+(b-a)r+c=0$ .

In this particular problem, $x^2y^{\prime\prime}-5xy^{\prime}+9y=0$ has the characteristic equation $r^2-6r+9=0$ , which yields repeated roots.

So once you have $y_1(x)$ , you suppose $y_2(x)=u(x)y_1(x)$ . It turns out that once you write the ODE in the form $y^{\prime\prime}+P(x)y^{\prime}+Q(x)y=0$ and substitute $y_2(x)=u(x)y_1(x)$ into it, we see that $u(x)=\displaystyle\int\left[\frac{e^{-\int P(x)dx}}{[y_1(x)]^2}\right]dx$ (this is the reduction of order technique as seen in my tutorial here.)

Then at this point, we see that $y_2(x)=\left(\displaystyle\int\left[\frac{e^{-\int P(x)dx}}{[y_1(x)]^2}\right]dx\right)y_1(x)$ is the second solution.

Can you continue with the problem?

**arbolis** · August 1st 2010, 05:44 PM

Thanks for the help, I will try. However isn't the characteristic equation $r^2-5r+9=0$ instead of $r^2-6r+9=0$ ? So that the roots aren't repeated. I should solve both problems anyway.

**Chris L T521** · August 1st 2010, 06:01 PM

Let $a=1$ , $b=-5$ and $c=9$ . In this particular case, it has to be $r^2+(-5-1)r+9=r^2-6r+9=0$ . Observe that $ax^2y^{\prime\prime}+bxy^{\prime}+cy=0$ and $ay^{\prime\prime}+by^{\prime}+cy=0$ have different characteristic equations since they assume different solutions (the former assumes solutions of the form $y=x^r$ and the latter assumes solutions of the form $y=e^{rx}$ )!

The former type of DE has the characteristic equation $ar^2+(b-a)r+c=0$ , where as the latter type has the equation $ar^2+br+c=0$ .

**arbolis** · August 1st 2010, 06:17 PM

I should have read better your first reply.
Ok, I get $y_1 (x)=x^3$ .
With $P(x)=-\frac{5}{x}$ , I reach that $u(x)=k \log |x| +C$ .
Thus the general solution would be of the form $y(x)=x^3 (c_1 \log |x| + c_2)+x^3=x^3(c_1 \log |x| + c_3)$ .
I hope it's right. And sorry, I didn't recognize it was a Cauchy-Euler equation.

**Chris L T521** · August 2nd 2010, 02:55 AM

Originally Posted by arbolis

I should have read better your first reply.
Ok, I get $y_1 (x)=x^3$ .
With $P(x)=-\frac{5}{x}$ , I reach that $u(x)=k \log |x| +C$ .
Thus the general solution would be of the form $y(x)=x^3 (c_1 \log |x| + c_2)+x^3=x^3(c_1 \log |x| + c_3)$ .
I hope it's right. And sorry, I didn't recognize it was a Cauchy-Euler equation.

No need to be sorry!

Its just something that is useful to know down the road. Now, when we go ahead and find $u(x)$ , we don't care much for the constants (they should drop out in the end anyways); so what where really after is that $u(x)=\log x$ (or $u(x)=\ln x$ , depending on what notation is used where you're at). Thus, we see that the general solution is $y=c_1x^3+c_2x^3\log x$ , which is what you have for your solution (but written in a slightly different way).

Hope that clarifies things!

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