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4.7 Cauchy-Euler Equations: Class Notes

4.7a

Definition

A Cauchy-Euler equation is a differential equation of the form ${\displaystyle a_n{x}^n\frac{d^{n}y}{d{x}^n}+a_{n-1}{x}^{n-1}\frac{d^{n-1}y}{d{x}^{n-1}}}$ ${\displaystyle +\cdots +a_{1}x\frac{dy}{dx}}$ $+a_{0}y=g(x)$. Earlier in the chapter we found solutions of the form $y=e^{mx}$. Here we will find solutions of the form $y=x^m$.

In each term, the power of $x$ matches the order of the derivative.

Solve the given differential equation.

Example

$$x^{2}y″+3xy\prime -4y=0$$

Solution

Let $y=x^m$. Then $y\prime =m{x}^{m-1}$ and $y″=m(m-1)x^{m-2}$. Substituting:

$$\begin{array}{cc}\hfill {\displaystyle x^{2}m(m-1)x^{m-2}+3xm{x}^{m-1}-4x^m}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle m(m-1)x^m+3m{x}^m-4x^m}& {\displaystyle =0}\hfill \end{array}$$

Divide by $x^m$ and complete the square:

$$\begin{array}{cc}\hfill {\displaystyle m(m-1)+3m-4}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle m^2+2m-4}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle m^2+2m+1}& {\displaystyle =4+1}\hfill \\ \hfill {\displaystyle (m+1{)}^2}& {\displaystyle =5}\hfill \\ \hfill {\displaystyle m}& {\displaystyle =-1\pm \sqrt{5}}\hfill \end{array}$$

In general, for $a{x}^{2}y″+bxy\prime +cy=0$, the auxiliary equation is

$$am(m-1)+bm+c=0,\text{that is,}a{m}^2-am+bm+c=0.$$

One of three results can stem from this quadratic equation:

Case I: $m_1\ne m_2$, both real.

$$y=c_1{x}^{m_1}+c_2{x}^{m_2}$$

Case II: $m_1=m_2=m$ (repeated). The second solution comes from reduction of order. In standard form the DE is

$${\displaystyle y″+\frac{b}{ax}y\prime +\frac{c}{a{x}^2}y=0,P(x)=\frac{b}{ax}.}$$

A repeated root means the discriminant is zero:

$${\displaystyle (-a+b{)}^2-4ac=0⟹m=\frac{a-b}{2a}⟹y_1=x^{(a-b)/(2a)}=x^m.}$$

So $x^{2m}=x^{(a-b)/a}=x^{1-b/a}$, and the reduction-of-order formula gives

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle y_2}& {\displaystyle =x^m\int \frac{e^{-\frac{b}{a}\int \frac{1}{x}dx}}{x^{(a-b)/a}}dx=x^m\int \frac{x^{-b/a}}{x^{1-b/a}}dx}\hfill \\ \hfill {\displaystyle }& {\displaystyle =x^m\int \frac{1}{x}dx=x^m\ln x.}\hfill \end{array}}$$

$$y_c=c_1{x}^m+c_2{x}^m\ln x$$

Case III: $m_1=\alpha +\beta i$, $m_2=\alpha -\beta i$.

$$y=x^{\alpha }[c_1\cos (\beta \ln x)+c_2\sin (\beta \ln x)]$$

(This is again due to Euler’s formula and the fact that $x^{\alpha }=e^{\alpha \ln x}$.)

Nonhomogeneous case: for $a{x}^{2}y″+bxy\prime +cy=g(x)$, divide by $a{x}^2$ to get the standard form $y″+P(x)y\prime +Q(x)y=f(x)$, then use variation of parameters.

4.7b

Example

$$4x^{2}y″+y=0$$

Solution

Let $y=x^m$. Then $y\prime =m{x}^{m-1}$ and $y″=m(m-1)x^{m-2}$, so

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle 4x^{2}m(m-1)x^{m-2}+x^m}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle x^m(4m^2-4m+1)}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle (2m-1{)}^2}& {\displaystyle =0⟹m=\frac{1}{2}\text{(repeated)}}\hfill \end{array}}$$

By Case II, $y_1=x^{1/2}$ and $y_2=x^{1/2}\ln x$.

Example

$$x^{2}y″-7xy\prime +41y=0$$

Solution

With $y=x^m$, the auxiliary equation is

$$\begin{array}{cc}\hfill {\displaystyle m(m-1)-7m+41}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle m^2-8m+41}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle m^2-8m+16}& {\displaystyle =-41+16}\hfill \\ \hfill {\displaystyle (m-4{)}^2}& {\displaystyle =-25⟹m=4\pm 5i}\hfill \end{array}$$

By Case III with $\alpha =4$ and $\beta =5$:

A Cauchy-Euler equation can be converted to an equation with constant coefficients using the substitution $x=e^t$ and the Chain Rule.

Rationale: The general solution to $ay″+by\prime +cy=0$ ($a,b,c$ constants) is

$$y=c_1{e}^{m_{1}x}+c_2{e}^{m_{2}x}.(*)$$

The general solution to $a{x}^{2}y″+bxy\prime +cy=0$ ($a,b,c$ constants) is $y=c_1{x}^{m_1}+c_2{x}^{m_2}$. But $x=e^{\ln x}⟹x^{m_1}=e^{m_1\ln x}$, so

$$y=c_1{e}^{m_1\ln x}+c_2{e}^{m_2\ln x}.$$

If we let $x=e^t$, we get $y=c_1{e}^{m_{1}t}+c_2{e}^{m_{2}t}$, resembling $(*)$.

Example

Use the substitution $x=e^t$ to transform the given Cauchy-Euler equation to a differential equation with constant coefficients. Solve the original equation by solving the new equation using techniques previously learned.

$$x^{2}y″-9xy\prime +25y=0$$

Solution

With $x=e^t$, we have $t=\ln x$, so $y(x)$ becomes $y(t(x))$ with $t(x)=\ln x$: $y$ depends on $t$, which depends on $x$. By the Chain Rule,

$${\displaystyle \frac{dy}{dx}=\frac{dy}{dt}·\frac{dt}{dx}=\frac{1}{x}\frac{dy}{dt}.}$$

Differentiate again with the Product Rule. The derivative ${\displaystyle \frac{d}{dx}\left(\frac{dy}{dt}\right)}$ also needs the Chain Rule: it is ${\displaystyle \frac{d^{2}y}{d{t}^2}·\frac{dt}{dx}=\frac{d^{2}y}{d{t}^2}·\frac{1}{x}}$.

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle \frac{d^{2}y}{d{x}^2}}& {\displaystyle =-\frac{1}{x^2}\frac{dy}{dt}+\frac{1}{x}\frac{d}{dx}\left(\frac{dy}{dt}\right)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =-\frac{1}{x^2}\frac{dy}{dt}+\frac{1}{x}(\frac{d^{2}y}{d{t}^2}·\frac{1}{x})=\frac{1}{x^2}(\frac{d^{2}y}{d{t}^2}-\frac{dy}{dt})}\hfill \end{array}}$$

Substitute into the DE; the powers of $x$ cancel:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle x^2·\frac{1}{x^2}(\frac{d^{2}y}{d{t}^2}-\frac{dy}{dt})-9x·\frac{1}{x}\frac{dy}{dt}+25y}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle \frac{d^{2}y}{d{t}^2}-10\frac{dy}{dt}+25y}& {\displaystyle =0}\hfill \end{array}}$$

The auxiliary equation is $m^2-10m+25=0$, or $(m-5{)}^2=0$, so $m=5$ (repeated):

$$\begin{array}{cc}\hfill {\displaystyle y}& {\displaystyle =c_1{e}^{5t}+c_{2}t{e}^{5t}}\hfill \\ \hfill {\displaystyle }& {\displaystyle =c_1{e}^{5\ln x}+c_2\ln x{e}^{5\ln x}}\hfill \end{array}$$