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

1. ## 2nd order DE (non linear I think but homogeneous)

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.

2. 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?

3. 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.

4. 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 $ar^2+br+c=0$.

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.

6. Originally Posted by arbolis
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 $y(x)=x^3 (c_1 \log |x| + c_2)+x^3=x^3(c_1 \log |x| + c_3)$.
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).