1
$\begingroup$

I have to find a fundamental set of solutions $\{y_1(x),y_2(x)\}$ for the equation $$x^2y''+3xy'+5y=0$$ I have followed all the proper steps, but so far, my answer registers as wrong. Can some help point out my mistake below? Thank you.

We substitute $y=x^m$, and so, $x^2(m(m-1)x^{m-2})+3xmx^{m-1}+5x^m=0$, which is simplified as: $$\left[m(m-1)+3m+5\right]x^m=0 \implies m^2+2m+5=0 \implies m=\frac{-2 \pm \sqrt{4-20}}{2}=\frac{-2 \pm 4i}{2}=-1 \pm 2i$$ We know that when $b^2-4ac<0$, the real basis for the null space is $\left[x^{\alpha}\cos(\beta \ln(x)),x^{\alpha}\sin(\beta \ln(x))\right]$, where $m_{1,2}=\alpha \pm i \beta$, and thus, the solutions are: $$y_1(x)=x^{-1}\cos(2\ln(x)) \\ y_2(x)=x^{-1}\sin(2\ln(x))$$

Edit: I have edited my error; the process is now correct.

$\endgroup$
2
  • 4
    $\begingroup$ Look at your roots $m^2 + 2m + 5 = 0$ again. What is the sqrt of $- 16$? $\endgroup$
    – Amzoti
    Feb 16, 2015 at 23:58
  • $\begingroup$ Oops, thanks! Can't believe I didn't see that this entire time. $\endgroup$
    – user208614
    Feb 17, 2015 at 0:02

1 Answer 1

0
$\begingroup$

I'll give another solution:

Write your equation as $$x^2\frac{{\rm d}^2y}{{\rm d}x^2}+3x\frac{{\rm d}y}{{\rm d}x}+5y = 0.$$Do the substituiton $x = e^t$. We have: $$t = \ln x \implies \frac{{\rm d}t}{{\rm d}x} = \frac{1}{x} = \frac{1}{e^t} = e^{-t}.$$ By the chain rule: $$\frac{{\rm d}y}{{\rm d}x} = \frac{{\rm d}y}{{\rm d}t}\frac{{\rm d}t}{{\rm d}x} = e^{-t}\frac{{\rm d}y}{{\rm d}t},$$ and: $$\begin{align} \frac{{\rm d}^2y}{{\rm d}x^2} &= \frac{{\rm d}}{{\rm d}x}\left(e^{-t}\frac{{\rm d}y}{{\rm d}t}\right) \\ &= \frac{{\rm d}}{{\rm d}x}(e^{-t})\frac{{\rm d}y}{{\rm d}t} +e^{-t} \frac{{\rm d}}{{\rm d}x}\left(\frac{{\rm d}y}{{\rm d}t}\right) \\ &= \frac{{\rm d}}{{\rm d}t}(e^{-t})\frac{{\rm d}t}{{\rm d}x}\frac{{\rm d}y}{{\rm d}t}+e^{-t}\frac{{\rm d}}{{\rm d}t}\left(\frac{{\rm d}y}{{\rm d}t}\right)\frac{{\rm d}t}{{\rm d}x} \\ &= -e^{-2t}\frac{{\rm d}y}{{\rm d}t}+e^{-2t}\frac{{\rm d}^2y}{{\rm d}t^2} \\ &= e^{-2t}\left(\frac{{\rm d}^2y}{{\rm d}t^2} - \frac{{\rm d}y}{{\rm d}t}\right).\end{align}$$ Plugging this into your equation yields: $$\begin{align} e^{2t}e^{-2t}\left(\frac{{\rm d}^2y}{{\rm d}t^2} - \frac{{\rm d}y}{{\rm d}t}\right)+3e^te^{-t}\frac{{\rm d}y}{{\rm d}t}+5y &= 0 \\ \frac{{\rm d}^2y}{{\rm d}t^2}+2\frac{{\rm d}y}{{\rm d}t} +5y &= 0,\end{align}$$ which has $r^2+2r+5 = 0$ as a characteristic equation. Solving: $$r = \frac{-2\pm \sqrt{4-4\cdot 1\cdot 5}}{2} = \frac{-2 \pm 4{\rm i}}{2} = -1\pm 2{\rm i}.$$

So the general solution is: $$\begin{align} y &= c_1 e^{-t}\cos\left(2t\right)+c_2e^{-t}\sin\left(2t\right) \\ y &= \frac{c_1}{x}\cos\left(2\ln x\right)+\frac{c_2}{x}\sin\left(2\ln x\right), \quad c_1,c_2 \in \Bbb R.\end{align}$$

$\endgroup$

You must log in to answer this question.