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My Problem is this given Cauchy-Euler equation: $$x^{3}y^{\prime\prime\prime}+2xy^{\prime}-2y=0$$

My Approach was: i can see this is a ordinary differential equation of third-order and i think its linear. I was told that this equation should have variable coefficients and that a stratgey to solve this equation should include a Transformation into a equation with contant coefficients. In this case, i think the coefficients are $x^3$ and $2x$.

So let $y=x^m$ with $m$ being a number. then: $$y=x^m\to y'=mx^{m-1},~~y'''=m(m-1)(m-2)x^{m-3}$$

applied into the equation gives:

$$x^3y'''+xy'-y=0\Longrightarrow x^{3}m(m-1)(m-2)x^{m-3}+2x\cdot m\cdot x^{m-1}-2x^{m}=0$$

$$m^{3}x^{m}-3m^{2}x^{m}+4m\cdot x^{m}-2x^{m}=0$$

$$(m^{3}-3m^{2}+4m-2)x^{m}=0$$

While $x\neq 0$: Thats why: $$m^{3}-3m^{2}+4m-2=0$$

So $m_1 = 1$ ... so now i am stuck. what can i do with the m, since i have it now? And do i Need only one $m$ or do i need all possible $m$ ?

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Besides to @Boris prime hint, you can divide $m^{3}-3m^{2}+4m-2$ by $m-1$ to see that $$m^{3}-3m^{2}+4m-2=(m-1)(m^2-2m+2)=(m-1)(m-\alpha)(m-\beta)$$ where $\alpha=1+i, ~\beta=1-i$. So $$y_c(x)=C_1x^1+C_2x^1\cos(\ln(x))+C_3x^1\sin(\ln(x))$$

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  • $\begingroup$ Oh Amzoti! I forgot it. Yes, there should have been an $x^1$. $\endgroup$
    – Mikasa
    Jun 25 '13 at 15:05
  • $\begingroup$ Excellent! $\oplus^1$ $\endgroup$
    – amWhy
    Jun 26 '13 at 0:08
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Hint: You can decrease the order of the equation setting $z=y/x$ and then $u=z'$.

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The three solutions of $m$ to your polynomial equation give the basis functions $y$ of the solution space to the homogenous equation. As you already have one $m$, you can obtain the other two solutions by polynomial division. Once you have those (which will be in the form $m = \alpha \pm \beta i$), the complete solution is give as the linear combination of the basis functions: $$ y = c_1 x^{m_1 = 1} + c_2 x^{\alpha} \cos(\beta \ln x) + c_3 x^{\alpha} \sin (\beta \ln x). $$

Edited: $m = \alpha + \beta i $ became $ m = \alpha \pm \beta i$ to emphasize that the two are conjugate roots.

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