# Fourth Order Cauchy-Euler Differential Equation (Repeated Complex Roots)

Now I'm studying differential equations on the Cauchy-Euler equation topic. I was just wondering how to deal with repeated complex roots in Euler-Cauchy equation. My textbook never says about this, so I tried to search in different textbooks, but seems most textbooks don't mention about this. I just came across this question in my mind.

For example, if the equation is $$x^4y'''' +10x^3y''' + 27x^2y'' +21xy' +4y = f(x)$$

My approach is going to be the same everytime; that is, finding the fundamental set of solutions and then particular solution. I first solve the associated homogeneous differential equation by guessing the solution in the form $y=x^m$, and plugging this into the equation which gives $$m(m-1)(m-2)(m-3)+10m(m-1)(m-2)+27m(m-1)+21m+4=0$$ $$m^4 + 4m^3 + 8m^2 + 8m + 4=0$$ $$(m^2 +2m +2)^2$$ $$m=1\pm i$$ So my general solution should be in the form: $$y(x) = [c_1x^{-1}\cos(\ln(x)) + c_2x^{-1}\sin(\ln(x)) + c_3y_3 + c_4y_4] + y_p$$ where $y_3$ and $y_4$ are other members in the fundamental set of solutions and $y_p$ is a particular solution which depends on the input $f(x)$.

My question is how can I solve for the rest of the fundamental set of solutions. Is there any nice formula for constructing the general solution of repeated complex roots as it has in the case of repeated real root?

(In the case of repeated real roots, it is said that if $m_1$ is a root of multiplicity $k$, then it can be shown that $$x^{m_1}, x^{m_1} \ln(x), ..., x^{m_1}( \ln(x))^{k-1}$$ are k linearly independent solutions. Yet I don't even know how to prove this.)

Any help or any suggestion would be greatly appreciated.

I have read the answer already, but just still wondering what is the motivation or the explanation behind this statement: In the case of homogeneous linear equation with constant coefficients ,when ${m_1}$ is a root of multuplicity $k$ of an nth-degree auxiliary equation, it can be shown that the linearly independent solution are $$e^{{m_1}x} , xe^{{m_1}x} , x^2 e^{{m_1}x} , ..., x^{k-1} e^{{m_1}x}$$

This comes from the transformation $u=\ln x$ or $x=e^u$. Then \begin{align}y^{\prime} & = \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=\frac1x\frac{dy}{du}\\ y^{\prime\prime} & =-\frac1{x^2}\frac{dy}{du}+\frac1{x^2}\frac{d^2y}{du^2}\\ y^{\prime\prime\prime} & =\frac2{x^3}\frac{dy}{du}-\frac3{x^3}\frac{d^2y}{du^2}+\frac1{x^3}\frac{d^3y}{du^3}\\ y^{(4)} & =-\frac6{x^4}\frac{dy}{du}+\frac{11}{x^4}\frac{d^2y}{du^2}-\frac6{x^4}\frac{d^3y}{du^3}+\frac1{x^4}\frac{d^4y}{du^4}\end{align} Substituting all that into the original homogeneous differential equation, $$\frac{d^4y_h}{du^4}+4\frac{d^3y_h}{du^3}+8\frac{d^2y_h}{du^2}+8\frac{dy_h}{du}+4y_h=0$$ We attempt to solve this with the trial function $y_h=u^{ru}$ and this works if $$r^4+4r^3+8r^2+8r+4=(r^2+2r+2)^2=0$$ Since these are double roots, the general solution to the homogeneous differential equation is \begin{align}y_h & =(c_1+c_2u)e^{-u}\cos u+(c_3+c_4u)e^{-u}\sin u\\ & =(c_1+c_2\ln x)x^{-1}\cos(\ln x)+(c_3+c_4\ln x)x^{-1}\sin(\ln x)\end{align} So the answer is that it's the same way you would obtain the general solution to a linear homogeneous differential equation with constant coefficients, just in terms of the transformed independent variable $u=\ln x$.
• Thanks, but could you give me some explanation for the motivation behind why we need to add another solution that results from multiplying the solution with $u$ when ${c_1}e^{-u} \cos(u)$ is a repeated root? Apr 9, 2016 at 6:26
• I have heard that when ${m_1}$ is a root of multuplicity $k$ of an nth-degree auxiliary equation, it can be shown that the linearly independent solution are $$e^{{m_1}x} , xe^{{m_1}x} , x^2 e^{{m_1}x} , ..., x^{k-1} e^{{m_1}x}$$ but just wondering what is the motivation or the explanation behind this? Apr 9, 2016 at 6:33
• Think about it in terms of differential operators, where $Df(x)=f^{\prime}(x)$. Then if $r$ is a root of the characteristic equation $(D-r)e^r=re^r-re^r=0$ so $e^r$ is a solution to the homogeneous equation. Then given a solution $f(x)$ to the homogeneous equation belonging to r, $$(D-r)(g(x)f(x))=g^{\prime}(x)f(x)+g(x)f^{\prime}(x)-rg(x)f(x)=g^{\prime}(x)f(x)+g(x)(D-r)f(x)=g^{\prime}(x)f(x)$$, so the operator $(D-r)$ has the effect of differentiating $g(x)$. A polynomial will be reduced to $0$ after a finite number of differentiations. Apr 9, 2016 at 6:44