0
$\begingroup$

The function $y_1=x^\frac{1}{2}$ is a solution of $4x^2y''+y=0$. What is the general solution of the linear homogenous differential equation on the interval $(0,\infty)$.

Can someone help me out. I am preparing for my test and don't know how to do this.

$\endgroup$
0

2 Answers 2

2
$\begingroup$

We have \begin{align} 4x^2y''\left(x\right)+y\left(x\right)=0.\tag{1} \end{align} This is a classic example of a Cauchy-Euler equidimensional equation. Substitute $y\left(x\right)=x^r$. Then \begin{align} y\left(x\right)=x^r,\\ y'\left(x\right)=rx^{r}x^{-1},\\ y''\left(x\right)=r\left(r-1\right)x^{r}x^{-2}. \end{align} This gives us \begin{align} 0&=4x^2\left(r\left(r-1\right)x^{r}x^{-2}\right)+\left(x^{r}\right)\\ &=4r\left(r-1\right)x^r+x^r\\ &=4r\left(r-1\right)+1.\tag{2} \end{align} Now solve this characteristic equation for $r$: \begin{align} r&=\frac{4\pm \sqrt{16-4\left(4\right)\left(1\right)}}{8}\\ &=\frac{4\pm 0}{8}=\frac{1}{2}.\tag{3} \end{align} Therefore we have $y\left(x\right)=C_0 x^{1/2}$ as a general solution, but one of repeated roots. To find the second solution you make the observation that $\frac{\partial }{\partial r}\left(x^r\right)=x^r\log\left(x\right)$. Therefore, the most "general" solution is \begin{align} y\left(x\right)=C_0x^{1/2}+C_1x^{1/2}\log\left(x\right),\;\;x\gt 0.\tag{4} \end{align}

$\endgroup$
4
  • $\begingroup$ ?? $\frac{\partial}{\partial x} x^r = r x^{r-1}$. I think you know what you were doing but the computation you wrote at the end is wrong. $\endgroup$ Mar 31, 2015 at 3:08
  • $\begingroup$ @PatrickDaSilva Ahh you are right. Thank you for catching my mistake. I was looking it over because I wasn't sure you meant my $\frac{d^n y}{dx^n}$'s or $\frac{\partial }{\partial x}\left(x^r\right)$, which was indeed wrong. $\endgroup$
    – bjd2385
    Mar 31, 2015 at 3:14
  • $\begingroup$ Can you fix it? Method of variation of coefficients maybe? I haven't done this in a while. $\endgroup$ Mar 31, 2015 at 10:40
  • $\begingroup$ @PatrickDaSilva I believe the coefficients are found via initial conditions, but they have not given me any. $\endgroup$
    – bjd2385
    Apr 2, 2015 at 14:12
-3
$\begingroup$

Hint: You can use Frobenius Method.

$\endgroup$
3
  • $\begingroup$ Totally irrelevant. $\endgroup$
    – user147263
    Mar 31, 2015 at 4:09
  • 1
    $\begingroup$ @Woodface: What do you mean? Of course it is relevant! $\endgroup$
    – science
    Mar 31, 2015 at 4:10
  • 1
    $\begingroup$ Would you care explaining what's the downvote for? $\endgroup$
    – science
    Mar 31, 2015 at 4:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .