Question

Find the general solution of (x+1)^2*y"+3(x+1)y'+0.75y=0 that is valid in any interval not including the singular point. (I did substitution of y=x^r, y'=rx^r-1, y"=r(r-1)x^r-2. But what can you factor out once you substitute those in?)

Answer 1

should tag expert one. you go ahead me. I didn't study singular point yet
@SithsAndGiggles

Answer 2

Have you tried \(y=(x+1)^r\) as a substitution? I think you'll have more luck.

Answer 3

Let me try, please wait a minute, don't leave.

Answer 4

This is what I got: (x+1)^2*(r(r-1)(x+1)^(r-2)+(x+1)^(r-1))+3(x+1)(r(x+1)^(r-1))+0.75(x+1)^r=0 but what can I factor out from here?

Answer 5

Substituting into the ODE,
\[(x+1)^2\left[r(r-1)(x+1)^{r-2}\right]+3(x+1)\left[r(x+1)^{r-1}\right]+\frac{3}{4}\left[(x+1)^r\right]=0\\
r(r-1)\color{blue}{(x+1)^r}+3r\color{blue}{(x+1)^r}+\frac{3}{4}\color{blue}{(x+1)^r}=0\]
A common factor! Since you're ignoring the singular point \((x=-1)\), you have \((x+1)^r\not=0\), so you can divide both sides by it:
\[r(r-1)+3r+\frac{3}{4}=0\\
r^2+2r+\frac{3}{4}=0\\
r=\cdots\]

Answer 6

Thank you.

Answer 7

Please wait a moment, let me work it out.

Answer 8

So I got r=1+1/2, 1-1/2, -1+1/2, and -1-1/2. How will the formula of this general solution be look like?

Answer 9

Your solutions for \(r\) aren't correct:
\[r=\frac{-2\pm\sqrt{4-4\times\frac{3}{4}}}{2}=-\frac{3}{2},-\frac{1}{2}\]
So now you have solutions, \(y_1=(x+1)^{r_1}\) and \(y_2=(x+1)^{r_2}\). The general solution to this ODE is given by \(y(x)=C_1y_1+C_2y_2\).

Answer 10

... where \(r_1=-\frac{3}{2}\) and \(r_2=-\frac{1}{2}\).

Answer 11

At the beginning, we tried y =(x+1)^r how can we know that form?

Answer 12

And the answer involves absolute value of x+1. How do I get that?

Answer 13

@Loser66, that's the kind of guess you make for this type of equation.

Given \(x^ny^{(n)}+x^{n-1}y^{(n-1)}+\cdots+xy'+y=0\), you'd use \(y=x^r\) as a guess. I forget the name, but it's named after some mathematician.

Answer 14

@Idealist, I'm not so sure about that. WA seems to agree with me: http://www.wolframalpha.com/input/?i=%28x%2B1%29%5E2*y%27%27%2B3%28x%2B1%29y%27%2B.75y%3D0

Answer 15

@Loser66, they're called Euler equations: http://en.wikipedia.org/wiki/Cauchy%E2%80%93Euler_equation

Answer 16

@Idealist I can answer that question, since \[y_1= (x+1)^{-\frac{3}{2}}\] so the form (x+1) under the squaroot, should be absolute

Answer 17

@SithsAndGiggles I got it. if the coefficient is x only , so we "guess" x ^r . In this case, it's (x+1) , so we "guess" it's (x+1)^r, right?

Answer 18

I see, thank you guys so much for the time and help. Have a good day.

Answer 19

@Loser66, yep, if the equation contains \((x-c)\) instead of just \(x\), you would guess \(y=(x-c)^r\).

@Idealist, you're welcome!

Answer 20

Thanks a ton. I didn't study this part yet, but learn a lot from you.

Answer 21

You're welcome too!