
# 6.5: Euler Equations

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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

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