6.5: Euler Equations

In this section, we will investigate the solutions of the most simple type of differential equations with regular singular points

\[ x^2y'' + axy' + by = 0 \]

We can immediately see that 0 is a regular singular point of the differential equation since

\[ x\; p(x) = a \;\;\; \text{and } \;\;\; x^2\; q(x) = b\]

To solve the differential equation we assume that a solution is of the form

\[ y = x^r \]

Now we proceed in a similar way to how we tackled second order differential equations with constant coefficients. We take two derivatives.

\[ y' = r \; x^{r-1} \]

\[ y'' = r(r-1) \; x^{r-2} \]

Next plug these into the original differential equation

\[\begin{align} x^2 r (r - 1)x^{r-2} + ax r x^{r-1} + b x^r &= 0 \\ r (r - 1)x^r + ar x^r + b x^r &= 0 &\text{Multiplying the exponents} \\ r (r - 1) + ar + b &= 0 &\text{Dividing by $x^r$} \\ r^2 + (a - 1) r + b &= 0. \end{align}\]

We define

\[ F(r) = r^2 + (a-1) r + b. \]

This is a quadratic in \(r \). We will see as with second order homogeneous linear differential equations with constant coefficients, there will be three cases.

Example \(\PageIndex{1}\): Distinct Roots

Solve

\[ x^2y'' + 5xy' + 3y = 0. \]

Solution

Let

\[ y = x^r \]

We take two derivatives.

\[ y' = r x ^{r-1} \]

\[ y'' = r(r-1)x^{r-2}. \]

Next plug these into the original differential equation

\[\begin{align} x^2 r ( r-1) x ^{r-2} + 5x\ ; rx^{r -1} + 3x^r &= 0 \\ r (r - 1)x^r + 5r x^r + 3x^r &= 0 &\text{Multiplying the exponents} \\ r (r - 1) + 5r + 3 &= 0 &\text{Dividing by $x^r$} \\ r^2 + 4 r + 3 &= 0 \\ (r + 3)(r + 1) &= 0 \end{align}\]

\[ r = -3 \;\;\; \text{or} \;\;\; r = -1. \]

The general solution is

\[ y = c_1x ^{-3} + c_2x ^{-1} .\]

Example \(\PageIndex{2}\)

Solve

\[ x^2y'' + 7xy' + 9y = 0 .\]

Solution

Let

\[ y = x^r .\]

We have

\[ F(r) = r^2 + 6r + 9 = (r + 3)^2 \]

which has the repeated root

\[ r = -3 .\]

Hence a solution is

\[ y_1 = x^{ -3} . \]

This gives us one solution and we could get another solution by reduction of order. However, there is a more clever way by noting that both \( F(r) \) and \( F'(r) \) are zero at \(r = -3 \).

Notice also that the partial derivative

\[ (x^r)_r = x^r \ln x .\]

We have

\[ L_r(x^r) = [x^rF(r)]_r \]

or

\[ L(x^r \ln r) = F(r) x^r \ln x + x^r F_r(r) = (r + 3)^2 x^r \ln x +2x^r (r + 3). \]

Now plug in \(r = -3\) to get

\[ L(x ^{-3} \ln x) = 0 .\]

Hence

\[ y_2 = x^{ -3} \ln x. \]

The general solution is

\[ y = c_1x ^{-3} + c_2x^{ -3} \ln x. \]

Example \(\PageIndex{3}\)

Solve

\[ x^2y'' + 5xy' + 8y = 0 . \]

Solution

Let

\[ y = x^r . \]

We have

\[ F(r) = r^2 + 4r + 8 . \]

which has complex roots

\[ r = 2 + 4i \;\;\; \text{ and } \;\;\; r = 2 - 4i \]

We get the solutions

\[ y_1 = x^{2 + 4i} \;\;\; \text{and} \;\;\; y_2 = x^{2 - 4i} \]

As with constant coefficients, we would like to express the solution without complex numbers. We have

\[ x ^{-2 + 2i} = e^{(-2 + 2i)\ln x} = x^{-2}e^{2\ln x i} = x^{-2}[\cos(2\ln x) + i \;\sin(2\ln x)]. \]

Similarly

\[ x^{ -2 - 2i} = e^{(-2 - 2i)\ln x} = x^{ -2}e^{-2\ln x i} = x^{ -2}[\cos(2\ln x) - i\; \sin(2\ln x)] . \]

By playing with constants we get the two solutions

\[ y_1 = x^{ -2} \cos(2\ln x) \;\;\; \text{and} \;\;\; y_2 = x ^{-2} \sin(2\ln x) \]

The general solution is

\[ y = c_1x ^{-2} \cos(2\ln x) + c_2x ^{-2} \sin(2\ln x) = x^{ -2}[c_1\cos(2\ln x) + c_2\sin(2\ln x)] \]

In summary, we have the following theorem.

Theorem: Solutions to Euler Equations

Let

\[ x^2y'' + axy' + by = 0 \]

and let

\[ F(r) = r^2 + (a - 1)r + b \]

have roots \( r_1 \) and \( r_2 \).

• Case 1: If \( r_1 \) and \( r_2 \) are real and distinct, then the general solution is

\[ y = c_1x^{r_1} + c_2x^{r_2}. \]

• Case 2: If \(r_1 = r_2 = r \) then the general solution is

\[ y = c_1x^r + c_2x^r \ln x. \]

• Case 3: If \( r_1 = l + mi \) and \( r_2 = l - mi \) then the general solution is

\[ y = x^l (c_1 \cos(m\; \ln x) + c_2 \sin(m\; \ln x)). \]

Contributors

• Larry Green (Lake Tahoe Community College)

• Integrated by Justin Marshall.