# 12.4: Cauchy-Euler Equations


Another class of solvable linear differential equations that is of interest are the Cauchy-Euler type of equations, also referred to in some books as Euler’s equation. These are given by $a x^{2} y^{\prime \prime}(x)+b x y^{\prime}(x)+c y(x)=0 .\label{eq:1}$ Note that in such equations the power of $$x$$ in each of the coefficients matches the order of the derivative in that term. These equations are solved in a manner similar to the constant coefficient equations.

One begins by making the guess $$y(x)=x^{r}$$. Inserting this function and its derivatives, $y^{\prime}(x)=r x^{r-1}, \quad y^{\prime \prime}(x)=r(r-1) x^{r-2},\nonumber$ into Equation $$\eqref{eq:1}$$, we have $[a r(r-1)+b r+c] x^{r}=0 .\nonumber$ Since this has to be true for all $$x$$ in the problem domain, we obtain the characteristic equation $a r(r-1)+b r+c=0 .\label{eq:2}$

### Note

The solutions of Cauchy-Euler equations can be found using the characteristic equation $$\operatorname{ar}(r-1)+b r+c=0$$.

Just like the constant coefficient differential equation, we have a quadratic equation and the nature of the roots again leads to three classes of solutions. If there are two real, distinct roots, then the general solution takes the form $$y(x)=c_{1} x^{r_{1}}+c_{2} x^{r_{2}} .$$

### Note

For two real, distinct roots, the general solution takes the form $y(x)=c_{1} x^{r_{1}}+c_{2} x^{r_{2}} .\nonumber$

### Example $$\PageIndex{1}$$

Find the general solution: $$x^{2} y^{\prime \prime}+5 x y^{\prime}+12 y=0$$.

###### Solution

As with the constant coefficient equations, we begin by writing down the characteristic equation. Doing a simple computation, \begin{align} 0 &=r(r-1)+5 r+12\nonumber \\ &=r^{2}+4 r+12\nonumber \\ &=(r+2)^{2}+8,\nonumber \\ -8 &=(r+2)^{2},\label{eq:3} \end{align} one determines the roots are $$r=-2 \pm 2 \sqrt{2}$$. Therefore, the general solution is $$y(x)=\left[c_{1} \cos (2 \sqrt{2} \ln |x|)+c_{2} \sin (2 \sqrt{2} \ln |x|)\right] x^{-2}$$

Deriving the solution for Case 2 for the Cauchy-Euler equations works in the same way as the second for constant coefficient equations, but it is a bit messier. First note that for the real root, $$r=r_{1}$$, the characteristic equation has to factor as $$\left(r-r_{1}\right)^{2}=0$$. Expanding, we have $r^{2}-2 r_{1} r+r_{1}^{2}=0 \text {. }\nonumber$ The general characteristic equation is $\operatorname{ar}(r-1)+b r+c=0 .\nonumber$ Dividing this equation by $$a$$ and rewriting, we have $r^{2}+\left(\frac{b}{a}-1\right) r+\frac{c}{a}=0 \text {. }\nonumber$ Comparing equations, we find $\frac{b}{a}=1-2 r_{1}, \quad \frac{c}{a}=r_{1}^{2} .\nonumber$ So, the Cauchy-Euler equation for this case can be written in the form $x^{2} y^{\prime \prime}+\left(1-2 r_{1}\right) x y^{\prime}+r_{1}^{2} y=0 .\nonumber$

Now we seek the second linearly independent solution in the form $$y_{2}(x)=$$ $$v(x) x^{r_{1}}$$. We first list this function and its derivatives, \begin{align} &y_{2}(x)=v x^{r_{1}},\nonumber \\ &y_{2}^{\prime}(x)=\left(x v^{\prime}+r_{1} v\right) x^{r_{1}-1},\nonumber \\ &y_{2}^{\prime \prime}(x)=\left(x^{2} v^{\prime \prime}+2 r_{1} x v^{\prime}+r_{1}\left(r_{1}-1\right) v\right) x^{r_{1}-2} .\label{eq:4} \end{align} Inserting these forms into the differential equation, we have \begin{align} 0 &=x^{2} y^{\prime \prime}+\left(1-2 r_{1}\right) x y^{\prime}+r_{1}^{2} y\nonumber \\ &=\left(x v^{\prime \prime}+v^{\prime}\right) x^{r_{1}+1} .\label{eq:5} \end{align} Thus, we need to solve the equation $x v^{\prime \prime}+v^{\prime}=0,\nonumber$ or $\frac{v^{\prime \prime}}{v^{\prime}}=-\frac{1}{x}\nonumber$ Integrating, we have $\ln \left|v^{\prime}\right|=-\ln |x|+C \text {, }\nonumber$ where $$A=\pm e^{C}$$ absorbs $$C$$ and the signs from the absolute values. Exponentiating, we obtain one last differential equation to solve, $v^{\prime}=\frac{A}{x} \text {. }\nonumber$ Thus, $v(x)=A \ln |x|+k .\nonumber$ So, we have found that the second linearly independent equation can be written as $y_{2}(x)=x^{r_{1}} \ln |x|.\nonumber$ Therefore, the general solution is found as $$y(x)=\left(c_{1}+c_{2} \ln |x|\right) x^{r}$$.

### Note

For one root, $$r_{1}=r_{2}=r$$, the general solution is of the form $y(x)=\left(c_{1}+c_{2} \ln |x|\right) x^{r} .\nonumber$

### Example $$\PageIndex{2}$$

Solve the initial value problem: $$t^{2} y^{\prime \prime}+3 t y^{\prime}+y=0$$, with the initial conditions $$y(1)=0, y^{\prime}(1)=1$$.

###### Solution

For this example the characteristic equation takes the form $r(r-1)+3 r+1=0,\nonumber$ or $r^{2}+2 r+1=0 .\nonumber$ There is only one real root, $$r=-1$$. Therefore, the general solution is $y(t)=\left(c_{1}+c_{2} \ln |t|\right) t^{-1} .\nonumber$

However, this problem is an initial value problem. At $$t=1$$ we know the values of $$y$$ and $$y^{\prime}$$. Using the general solution, we first have that $0=y(1)=c_{1} .\nonumber$ Thus, we have so far that $$y(t)=c_{2} \ln |t| t^{-1}$$. Now, using the second condition and $y^{\prime}(t)=c_{2}(1-\ln |t|) t^{-2},\nonumber$ we have $1=y(1)=c_{2} .\nonumber$ Therefore, the solution of the initial value problem is $$y(t)=\ln |t| t^{-1}$$.

We now turn to the case of complex conjugate roots, $$r=\alpha \pm i \beta$$. When dealing with the Cauchy-Euler equations, we have solutions of the form $$y(x)=x^{\alpha+i \beta}$$. The key to obtaining real solutions is to first rewrite $$x^{y}$$ : $x^{y}=e^{\ln x^{y}}=e^{y \ln x} .\nonumber$ Thus, a power can be written as an exponential and the solution can be written as $y(x)=x^{\alpha+i \beta}=x^{\alpha} e^{i \beta \ln x}, \quad x>0 .\nonumber$ Recalling that $e^{i \beta \ln x}=\cos (\beta \ln |x|)+i \sin (\beta \ln |x|),\nonumber$ we can now find two real, linearly independent solutions, $$x^{\alpha} \cos (\beta \ln |x|)$$ and $$x^{\alpha} \sin (\beta \ln |x|)$$ following the same steps as earlier for the constant coefficient case. This gives the general solution as $y(x)=x^{\alpha}\left(c_{1} \cos (\beta \ln |x|)+c_{2} \sin (\beta \ln |x|)\right) .\nonumber$

### Note

For complex conjugate roots, $$r=\alpha \pm i \beta$$, the general solution takes the form $$y(x)=x^{\alpha}\left(c_{1} \cos (\beta \ln |x|)+c_{2} \sin (\beta \ln |x|)\right)$$.

### Example $$\PageIndex{3}$$

Solve: $$x^{2} y^{\prime \prime}-x y^{\prime}+5 y=0$$.

###### Solution

The characteristic equation takes the form $r(r-1)-r+5=0,\nonumber$ or $r^{2}-2 r+5=0\nonumber$ The roots of this equation are complex, $$r_{1,2}=1 \pm 2$$. Therefore, the general solution is $$y(x)=x\left(c_{1} \cos (2 \ln |x|)+c_{2} \sin (2 \ln |x|)\right)$$.

The three cases are summarized in the table below.

Table $$\PageIndex{1}$$
Classification of Roots of the Characteristic Equation for Cauchy-Euler Differential Equations
1. Real, distinct roots $$r_{1}, r_{2}$$. In this case the solutions corresponding to each root are linearly independent. Therefore, the general solution is simply $$y(x)=c_{1} x^{r_{1}}+c_{2} x^{r_{2}}$$.
2. Real, equal roots $$r_{1}=r_{2}=r$$. In this case the solutions corresponding to each root are linearly dependent. To find a second linearly independent solution, one uses the Method of Reduction of Order. This gives the second solution as $$x^{r} \ln |x|$$. Therefore, the general solution is found as $$y(x)=\left(c_{1}+c_{2} \ln |x|\right) x^{r}$$.
3. Complex conjugate roots $$r_{1}, r_{2}=\alpha \pm i \beta$$. In this case the solutions corresponding to each root are linearly independent. These complex exponentials can be rewritten in terms of trigonometric functions. Namely, one has that $$x^{\alpha} \cos (\beta \ln |x|)$$ and $$x^{\alpha} \sin (\beta \ln |x|)$$ are two $$\operatorname{lin}$$ early independent solutions. Therefore, the general solution becomes $$y(x)=x^{\alpha}\left(c_{1} \cos (\beta \ln |x|)+c_{2} \sin (\beta \ln |x|)\right)$$.

## Nonhomogeneous Cauchy-Euler Equations

We can also solve some nonhomogeneous Cauchy-Euler equations using the Method of Undetermined Coefficients or the Method of Variation of Parameters. We will demonstrate this with a couple of examples.

### Example $$\PageIndex{4}$$

Find the solution of $$x^{2} y^{\prime \prime}-x y^{\prime}-3 y=2 x^{2}$$.

###### Solution

First we find the solution of the homogeneous equation. The characteristic equation is $$r^{2}-2 r-3=0$$. So, the roots are $$r=-1,3$$ and the solution is $$y_{h}(x)=c_{1} x^{-1}+c_{2} x^{3}$$.

We next need a particular solution. Let’s guess $$y_{p}(x)=A x^{2}$$. Inserting the guess into the nonhomogeneous differential equation, we have \begin{align} 2 x^{2} &=x^{2} y^{\prime \prime}-x y^{\prime}-3 y=2 x^{2}\nonumber \\ &=2 A x^{2}-2 A x^{2}-3 A x^{2}\nonumber \\ &=-3 A x^{2} .\label{eq:6} \end{align} So, $$A=-2 / 3$$. Therefore, the general solution of the problem is $y(x)=c_{1} x^{-1}+c_{2} x^{3}-\frac{2}{3} x^{2} .\nonumber$

### Example $$\PageIndex{5}$$

Find the solution of $$x^{2} y^{\prime \prime}-x y^{\prime}-3 y=2 x^{3}$$.

###### Solution

In this case the nonhomogeneous term is a solution of the homogeneous problem, which we solved in the last example. So, we will need a modification of the method. We have a problem of the form $a x^{2} y^{\prime \prime}+b x y^{\prime}+c y=d x^{r},\nonumber$ where $$r$$ is a solution of ar $$(r-1)+b r+c=0$$. Let’s guess a solution of the form $$y=A x^{r} \ln x$$. Then one finds that the differential equation reduces to $$A x^{r}(2 a r-$$ $$a+b)=d x^{r}$$. [You should verify this for yourself.]

With this in mind, we can now solve the problem at hand. Let $$y_{p}=A x^{3} \ln x$$. Inserting into the equation, we obtain $$4 A x^{3}=2 x^{3}$$, or $$A=1 / 2$$. The general solution of the problem can now be written as $y(x)=c_{1} x^{-1}+c_{2} x^{3}+\frac{1}{2} x^{3} \ln x .\nonumber$

### Example $$\PageIndex{6}$$

Find the solution of $$x^{2} y^{\prime \prime}-x y^{\prime}-3 y=2 x^{3}$$ using Variation of Parameters.

###### Solution

As noted in the previous examples, the solution of the homogeneous problem has two linearly independent solutions, $$y_{1}(x)=x^{-1}$$ and $$y_{2}(x)=x^{3}$$. Assuming a particular solution of the form $$y_{p}(x)=c_{1}(x) y_{1}(x)+c_{2}(x) y_{2}(x)$$, we need to solve the system (12.3.23): \begin{align} c_{1}^{\prime}(x) x^{-1}+c_{2}^{\prime}(x) x^{3} &=0\nonumber \\ -c_{1}^{\prime}(x) x^{-2}+3 c_{2}^{\prime}(x) x^{2} &=\frac{2 x^{3}}{x^{2}}=2 x .\label{eq:7} \end{align}

From the first equation of the system we have $$c_{1}^{\prime}(x)=-x^{4} c_{2}^{\prime}(x)$$. Substituting this into the second equation gives $$c_{2}^{\prime}(x)=\frac{1}{2 x} .$$ So, $$c_{2}(x)=\frac{1}{2} \ln |x|$$ and, therefore, $$c_{1}(x)=\frac{1}{8} x^{4}$$. The particular solution is $y_{p}(x)=c_{1}(x) y_{1}(x)+c_{2}(x) y_{2}(x)=\frac{1}{8} x^{3}+\frac{1}{2} x^{3} \ln |x| .\nonumber$ Adding this to the homogeneous solution, we obtain the same solution as in the last example using the Method of Undetermined Coefficients. However, since $$\frac{1}{8} x^{3}$$ is a solution of the homogeneous problem, it can be absorbed into the first terms, leaving $y(x)=c_{1} x^{-1}+c_{2} x^{3}+\frac{1}{2} x^{3} \ln x .\nonumber$

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