How Can I Solve This Second Order Linear ODE Problem?

[danthatdude]

I'm having a lot of trouble with this problem. I'm also having a lot of trouble inputting it into LaTeX. I hope you can follow even though the markup isn't good.

I'm trying to find a formula for the general solution of $ax^2y''+bxy'+cy=0$ where $y=x^r\ln(x)$ when $(b-a)^2-4ac=0$;

using product rule,
i got

$y'=x^{r-1}+rx^{r-1}\ln(x)$
and $y"=(r-1)x^{r-2}+rx^{r-2}\ln(x)$

substituting y, y', and y" into the differential equation, I got

$a((r-1)+r+r(r-1)\ln(x))+b(1+r\ln(x)+c\ln(x)=0$

(Since every term had $x^r$ after distributing $x^2$, x into the parentheses, I simplified it to 1 by dividing the entire equation by $x^r$).

I end up with
$a(((r-1)(r)\ln(x))+(2r-1))+b(1+r\ln(x)+c\ln(x)=0$

$a(r^2\ln(x)-r\ln(x)+2r-1)+b+br\ln(x)+c\ln(x)=0$

I'm not able to simplify this quadratic in a form that gives the solution to the roots with $(b-a)^2-4ac$ under the radical. (I solved a problem similar to this with a simpler function for y).

Can anybody point out where I went wrong, and how to arrive at the solution step by step? I went over the algebra very carefully but I can't pinpoint where my algebra went wrong. I feel like I'm very close to the solution but I went wrong at some very elementary step.

I also tried doing the markup for the problem with LaTeX but it wasn't typesetting correctly -- I don't have a lot of experience with it. I hope you're able to follow.

 

[mathmari]

I'm having a lot of trouble with this problem. I'm also having a lot of trouble inputting it into LaTeX. I hope you can follow even though the markup isn't good.

I'm trying to find a formula for the general solution of $ax^2y''+bxy'+cy=0$ where $y=x^r\ln(x)$ when $(b-a)^2-4ac=0$;

using product rule,
i got

$y'=x^{r-1}+rx^{r-1}\ln(x)$
and $y"=(r-1)x^{r-2}+rx^{r-2}\ln(x)$

substituting y, y', and y" into the differential equation, I got

$a((r-1)+r+r(r-1)\ln(x))+b(1+r\ln(x)+c\ln(x)=0$

(Since every term had $x^r$ after distributing $x^2$, x into the parentheses, I simplified it to 1 by dividing the entire equation by $x^r$).

I end up with
$a(((r-1)(r)\ln(x))+(2r-1))+b(1+r\ln(x)+c\ln(x)=0$

$a(r^2\ln(x)-r\ln(x)+2r-1)+b+br\ln(x)+c\ln(x)=0$

I'm not able to simplify this quadratic in a form that gives the solution to the roots with $(b-a)^2-4ac$ under the radical. (I solved a problem similar to this with a simpler function for y).

Can anybody point out where I went wrong, and how to arrive at the solution step by step? I went over the algebra very carefully but I can't pinpoint where my algebra went wrong. I feel like I'm very close to the solution but I went wrong at some very elementary step.

I also tried doing the markup for the problem with LaTeX but it wasn't typesetting correctly -- I don't have a lot of experience with it. I hope you're able to follow.

When you derivate $y'=x^{r-1}+rx^{r-1}\ln(x)$ you want to derivate a product.
So, the second derivative of $y$ is the following:
$$y''=(x^{r-1}+rx^{r-1}\ln(x))'=(x^{r-1})'+(rx^{r-1}\ln(x))' \\ =(r-1)x^{r-2}+(rx^{r-1})' \ln(x)+rx^{r-1}(\ln(x))'=(r-1)x^{r-2}+r(r-1)x^{r-2} \ln(x)+rx^{r-1}\frac{1}{x} \\ =(r-1)x^{r-2}+r(r-1)x^{r-2} \ln(x)+rx^{r-2}$$

 

[danthatdude]

When you derivate $y'=x^{r-1}+rx^{r-1}\ln(x)$ you want to derivate a product.
So, the second derivative of $y$ is the following:
$$y''=(x^{r-1}+rx^{r-1}\ln(x))'=(x^{r-1})'+(rx^{r-1}\ln(x))' \\ =(r-1)x^{r-2}+(rx^{r-1})' \ln(x)+rx^{r-1}(\ln(x))'=(r-1)x^{r-2}+r(r-1)x^{r-2} \ln(x)+rx^{r-1}\frac{1}{x} \\ =(r-1)x^{r-2}+r(r-1)x^{r-2} \ln(x)+rx^{r-2}$$

oops, i left out that last term, yes I got what you have for the second derivative.

But the issue still remains when I substitute it into the expression

I end up with a[(r-1)+r+(r)(r-1)ln(x)]+b[r(ln(x))]+cln(x)=0

after simplifying. the natural logarithm term is what's making it hard to get it into the quadratic equation in terms of r that I need it to be to be able to factor it into

[(a-b)+sqrt((b-a)^2-4ac)]/2a

 

[mathmari]

oops, i left out that last term, yes I got what you have for the second derivative.

But the issue still remains when I substitute it into the expression

I end up with a[(r-1)+r+(r)(r-1)ln(x)]+b[r(ln(x))]+cln(x)=0

after simplifying. the natural logarithm term is what's making it hard to get it into the quadratic equation in terms of r that I need it to be to be able to factor it into

[(a-b)+sqrt((b-a)^2-4ac)]/2a

$y=x^r \ln{(x)}$
$y'=x^{r-1}+rx^{r-1}\ln{(x)}$
$y''=(r-1)x^{r-2}+r(r-1)x^{r-2}\ln{(x)}+rx^{r-2}$

$$ax^2y''+bxy'+cy=0 \\ \Rightarrow ax^2 \left ( (r-1)x^{r-2}+r(r-1)x^{r-2}\ln{(x)}+rx^{r-2} \right )+bx \left ( x^{r-1}+rx^{r-1}\ln{(x)} \right )+cx^r \ln{(x)}=0 \\ \Rightarrow ax^r \left ( (r-1)+r(r-1)\ln{(x)}+r\right )+bx^r \left ( 1+r\ln{(x)} \right )+cx^r \ln{(x)}=0 \\ \Rightarrow x^r \left ( a \left ( 2r-1+r(r-1)\ln{(x)}\right )+b \left ( 1+r\ln{(x)} \right )+c \ln{(x)} \right )=0 \\ \Rightarrow x^r \left ( a(2r-1)+ar(r-1)\ln{(x)}+ b+br\ln{(x)} +c \ln{(x)} \right )=0 \\ \Rightarrow x^r \left ( 2ar-a+(ar^2-ar)\ln{(x)}+ b+br\ln{(x)} +c \ln{(x)} \right )=0 \\ \Rightarrow x^r \left ( a \ln{(x)} r^2+(2a+(b-a) \ln{(x)})r+(c\ln{(x)}+(b-a))\right )=0 \\ \Rightarrow a \ln{(x)} r^2+(2a+(b-a) \ln{(x)})r+(c\ln{(x)}+(b-a))=0 $$

The discriminant is equal to $$ \Delta=(2a+(b-a) \ln{(x)})^2-4 \cdot a \ln{(x)} \cdot (c\ln{(x)}+(b-a)) \\ =4a^2+4a(b-a) \ln{(x)}+(b-a)^2 \ln^2{(x)}-4ac \ln^2{(x)}-4a(b-a)\ln{(x)} \\ = 4a^2+\left ( (b-a)^2 -4ac \right )\ln^2{(x)} $$ Therefore, the solutions are $$ r_{1,2}=\frac{-(2a+(b-a) \ln{(x)}) \pm \sqrt{4a^2+\left ( (b-a)^2 -4ac \right )\ln^2{(x)}}}{2a \ln{(x)}}$$

 

[danthatdude]

$y=x^r \ln{(x)}$
$y'=x^{r-1}+rx^{r-1}\ln{(x)}$
$y''=(r-1)x^{r-2}+r(r-1)x^{r-2}\ln{(x)}+rx^{r-2}$

$$ax^2y''+bxy'+cy=0 \\ \Rightarrow ax^2 \left ( (r-1)x^{r-2}+r(r-1)x^{r-2}\ln{(x)}+rx^{r-2} \right )+bx \left ( x^{r-1}+rx^{r-1}\ln{(x)} \right )+cx^r \ln{(x)}=0 \\ \Rightarrow ax^r \left ( (r-1)+r(r-1)\ln{(x)}+r\right )+bx^r \left ( 1+r\ln{(x)} \right )+cx^r \ln{(x)}=0 \\ \Rightarrow x^r \left ( a \left ( 2r-1+r(r-1)\ln{(x)}\right )+b \left ( 1+r\ln{(x)} \right )+c \ln{(x)} \right )=0 \\ \Rightarrow x^r \left ( a(2r-1)+ar(r-1)\ln{(x)}+ b+br\ln{(x)} +c \ln{(x)} \right )=0 \\ \Rightarrow x^r \left ( 2ar-a+(ar^2-ar)\ln{(x)}+ b+br\ln{(x)} +c \ln{(x)} \right )=0 \\ \Rightarrow x^r \left ( a \ln{(x)} r^2+(2a+(b-a) \ln{(x)})r+(c\ln{(x)}+(b-a))\right )=0 \\ \Rightarrow a \ln{(x)} r^2+(2a+(b-a) \ln{(x)})r+(c\ln{(x)}+(b-a))=0 $$

The discriminant is equal to $$ \Delta=(2a+(b-a) \ln{(x)})^2-4 \cdot a \ln{(x)} \cdot (c\ln{(x)}+(b-a)) \\ =4a^2+4a(b-a) \ln{(x)}+(b-a)^2 \ln^2{(x)}-4ac \ln^2{(x)}-4a(b-a)\ln{(x)} \\ = 4a^2+\left ( (b-a)^2 -4ac \right )\ln^2{(x)} $$ Therefore, the solutions are $$ r_{1,2}=\frac{-(2a+(b-a) \ln{(x)}) \pm \sqrt{4a^2+\left ( (b-a)^2 -4ac \right )\ln^2{(x)}}}{2a \ln{(x)}}$$

Thank you so much. It looks like I got up to the very last step before the discriminant...I just factored differently and canceled x^r in earlier steps. I think had I not been trying to find a way to algebraically cancel out ln(x) I could have had the solution by using the formula for roots of a quadratic. I thought I had to cancel out ln(x).

So it's ok to have a function in r for y=(x^r)ln(x)?

 

[mathmari]

So it's ok to have a function in r for y=(x^r)ln(x)?

Yes! The general solution will be of the form $$y(x)=\left ( c_1 x^{r_1}+c_2 x^{r_2} \right ) \ln{(x)}$$