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Mathematics LibreTexts

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 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 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 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

  • Integrated by Justin Marshall.