2.5: 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

\[a x^{2} y^{\prime \prime}(x)+b x y^{\prime}(x)+c y(x)=0 \label{2.95} \]

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,

\(y^{\prime}(x)=r x^{r-1}, \quad y^{\prime \prime}(x)=r(r-1) x^{r-2}\)

into Equation \(\PageIndex{1}\), we have

\([a r(r-1)+b r+c] x^{r}=0\)

Since this has to be true for all \(x\) in the problem domain, we obtain the characteristic equation

\[a r(r-1)+b r+c=0 \nonumber \]

The solutions of Cauchy-Euler equations can be found using this characteristic equation. 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

\[y(x)=c_{1} x^{r_{1}}+c_{2} x^{r_{2}} \nonumber\]

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,

\[ \begin{aligned} 0 &=r(r-1)+5 r+12 \\[4pt] &=r^{2}+4 r+12 \\[4pt] &=(r+2)^{2}+8, \\[4pt] -8 &=(r+2)^{2}, \end{aligned} \label{2.97} \]

one determines the roots are \(r=-2 \pm 2 \sqrt{2} i\). Therefore, the general solution is

\[y(x)=\left[c_{1} \cos (2 \sqrt{2} \ln |x|)+c_{2} \sin (2 \sqrt{2} \ln |x|)\right] x^{-2} \nonumber\]

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

\(r^{2}-2 r_{1} r+r_{1}^{2}=0\)

The general characteristic equation is

\(a r(r-1)+b r+c=0\)

Dividing this equation by \(a\) and rewriting, we have

\(r^{2}+\left(\dfrac{b}{a}-1\right) r+\dfrac{c}{a}=0\)

Comparing equations, we find

\(\dfrac{b}{a}=1-2 r_{1}, \quad \dfrac{c}{a}=r_{1}^{2}\)

So, the Cauchy-Euler equation for this case can be written in the form

\(x^{2} y^{\prime \prime}+\left(1-2 r_{1}\right) x y^{\prime}+r_{1}^{2} y=0\)

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,

\[\begin{equation} \begin{aligned} &y_{2}(x)=v x^{r_{1}} \\[4pt] &y_{2}^{\prime}(x)=\left(x v^{\prime}+r_{1} v\right) x^{r_{1}-1} \\[4pt] &y_{2}^{\prime \prime}(x)=\left(x^{2} v^{\prime \prime}+2 r_{1} x v^{\prime}+r_{1}\left(r_{1}-1\right) v\right) x^{r_{1}-2} \end{aligned}\end{equation}\label{2.98} \]

Inserting these forms into the differential equation, we have

\[ \begin{aligned} 0 &=x^{2} y^{\prime \prime}+\left(1-2 r_{1}\right) x y^{\prime}+r_{1}^{2} y \\[4pt] &=\left(x v^{\prime \prime}+v^{\prime}\right) x^{r_{1}+1} \end{aligned} \label{2.99} \]

Thus, we need to solve the equation

\(x v^{\prime \prime}+v^{\prime}=0\)

Or

\(\dfrac{v^{\prime \prime}}{v^{\prime}}=-\dfrac{1}{x}\)

Integrating, we have

\(\ln \left|v^{\prime}\right|=-\ln |x|+C\)

where \(A=\pm e^{C}\) absorbs \(C\) and the signs from the absolute values. Exponentiating, we obtain one last differential equation to solve,

\(v^{\prime}=\dfrac{A}{x}\)

Thus,

\(v(x)=A \ln |x|+k\)

So, we have found that the second linearly independent equation can be written as

\(y_{2}(x)=x^{r_{1}} \ln |x|.\)

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 \(y(x)=\left(c_{1}+c_{2} \ln |x|\right) x^{r}.\)

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

\(r(r-1)+3 r+1=0\)

or

\(r^{2}+2 r+1=0\)

There is only one real root, \(r=-1\). Therefore, the general solution is

\(y(t)=\left(c_{1}+c_{2} \ln |t|\right) t^{-1}\)

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

\(0=y(1)=c_{1}\)

Thus, we have so far that \(y(t)=c_{2} \ln |t| t^{-1}\).

Now, using the second condition and

\(y^{\prime}(t)=c_{2}(1-\ln |t|) t^{-2}\)

we have

\(1=y(1)=c_{2}\)

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

\[y(x)=x^{\alpha+i \beta}=x^{\alpha} e^{i \beta \ln x}, \quad x>0 \nonumber \]

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

Recalling that

\(e^{i \beta \ln x}=\cos (\beta \ln |x|)+i \sin (\beta \ln |x|)\),

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

\(y(x)=x^{\alpha}\left(c_{1} \cos (\beta \ln |x|)+c_{2} \sin (\beta \ln |x|)\right)\)

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, \(r_{1,2}=1 \pm 2 i\). 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 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 \[y(x)=c_{1} x^{r_{1}}+c_{2} x^{r_{2}}. \nonumber\]
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 \[y(x)=\left(c_{1}+c_{2} \ln |x|\right) x^{r}. \nonumber \]
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 linearly independent solutions. Therefore, the general solution becomes \[y(x)=x^{\alpha}\left(c_{1} \cos (\beta \ln |x|)+c_{2} \sin (\beta \ln |x|)\right). \nonumber\]

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

\[ \begin{aligned} 2 x^{2} &=x^{2} y^{\prime \prime}-x y^{\prime}-3 y=2 x^{2} \\[4pt] &=2 A x^{2}-2 A x^{2}-3 A x^{2} \\[4pt] &=-3 A x^{2} \end{aligned}\label{2.100} \]

So, \(A=-2 / 3\). Therefore, the general solution of the problem is

\(y(x)=c_{1} x^{-1}+c_{2} x^{3}-\dfrac{2}{3} x^{2}\)

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

\(a x^{2} y^{\prime \prime}+b x y^{\prime}+c y=d x^{r}\)

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

\(y(x)=c_{1} x^{-1}+c_{2} x^{3}+\dfrac{1}{2} x^{3} \ln x\)

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 \(2.4.25\):

\[ \begin{aligned} c_{1}^{\prime}(x) x^{-1}+c_{2}^{\prime}(x) x^{3} &=0 \\[4pt] -c_{1}^{\prime}(x) x^{-2}+3 c_{2}^{\prime}(x) x^{2} &=\dfrac{2 x^{3}}{x^{2}}=2 x . \end{aligned}\label{2.101} \]

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 \(c_{2}^{\prime}(x)=\dfrac{1}{2 x} .\) So, \(c_{2}(x)=\) \(\dfrac{1}{2} \ln |x|\) and, therefore, \(c_{1}(x)=\dfrac{1}{8} x^{4}\). The particular solution is

\(y_{p}(x)=c_{1}(x) y_{1}(x)+c_{2}(x) y_{2}(x)=\dfrac{1}{8} x^{3}+\dfrac{1}{2} x^{3} \ln |x|\)

Adding this to the homogeneous solution, we obtain the same solution as in the last example using the Method of Undetermined Coefficients. However, since \(\dfrac{1}{8} x^{3}\) is a solution of the homogeneous problem, it can be absorbed into the first terms, leaving

\(y(x)=c_{1} x^{-1}+c_{2} x^{3}+\dfrac{1}{2} x^{3} \ln x\)