12.4: Cauchy-Euler Equations
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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 \[a r(r-1)+b r+c=0 .\label{eq:2}\]
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 \(y(x)=c_{1} x^{r_{1}}+c_{2} x^{r_{2}} .\)
For two real, distinct roots, the general solution takes the form \[y(x)=c_{1} x^{r_{1}}+c_{2} x^{r_{2}} .\nonumber \]
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 \(y(x)=\left(c_{1}+c_{2} \ln |x|\right) x^{r}\).
For one root, \(r_{1}=r_{2}=r\), the general solution is of the form \[y(x)=\left(c_{1}+c_{2} \ln |x|\right) x^{r} .\nonumber \]
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 \[y(x)=x^{\alpha}\left(c_{1} \cos (\beta \ln |x|)+c_{2} \sin (\beta \ln |x|)\right) .\nonumber \]
For complex conjugate roots, \(r=\alpha \pm i \beta\), the general solution takes the form \(y(x)=x^{\alpha}\left(c_{1} \cos (\beta \ln |x|)+c_{2} \sin (\beta \ln |x|)\right)\).
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.
| 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.
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 \[y(x)=c_{1} x^{-1}+c_{2} x^{3}-\frac{2}{3} x^{2} .\nonumber \]
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 \[y(x)=c_{1} x^{-1}+c_{2} x^{3}+\frac{1}{2} x^{3} \ln x .\nonumber \]
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 \[y(x)=c_{1} x^{-1}+c_{2} x^{3}+\frac{1}{2} x^{3} \ln x .\nonumber \]