6.5: Euler Equations
- Page ID
- 389
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)In this section, we will investigate the solutions of the most simple type of differential equations with regular singular points
\[ x^2y'' + axy' + by = 0 \nonumber \]
We can immediately see that 0 is a regular singular point of the differential equation since
\[ x\; p(x) = a \;\;\; \text{and } \;\;\; x^2\; q(x) = b\nonumber \]
To solve the differential equation we assume that a solution is of the form
\[ y = x^r \nonumber \]
Now we proceed in a similar way to how we tackled second order differential equations with constant coefficients. We take two derivatives.
\[ y' = r \; x^{r-1} \nonumber \]
\[ y'' = r(r-1) \; x^{r-2} \nonumber \]
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
\[ F(r) = r^2 + (a-1) r + b. \nonumber \]
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.
Solve
\[ x^2y'' + 5xy' + 3y = 0. \nonumber \]
Solution
Let
\[ y = x^r \nonumber \]
We take two derivatives.
\[ y' = r x ^{r-1} \nonumber \]
\[ y'' = r(r-1)x^{r-2}. \nonumber \]
Next plug these into the original differential equation
\[\begin{align*} x^2 r ( r-1) x ^{r-2} + 5x\ ; rx^{r -1} + 3x^r &= 0 \\ r (r - 1)x^r + 5r x^r + 3x^r &= 0 &\text{Multiplying the exponents} \\ r (r - 1) + 5r + 3 &= 0 &\text{Dividing by $x^r$} \\ r^2 + 4 r + 3 &= 0 \\ (r + 3)(r + 1) &= 0 \end{align*}\nonumber \]
\[ r = -3 \;\;\; \text{or} \;\;\; r = -1. \nonumber \]
The general solution is
\[ y = c_1x ^{-3} + c_2x ^{-1} .\nonumber \]
Solve
\[ x^2y'' + 7xy' + 9y = 0 .\nonumber \]
Solution
Let
\[ y = x^r .\nonumber \]
We have
\[ F(r) = r^2 + 6r + 9 = (r + 3)^2 \nonumber \]
which has the repeated root
\[ r = -3 .\nonumber \]
Hence a solution is
\[ y_1 = x^{ -3} . \nonumber \]
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
\[ (x^r)_r = x^r \ln x .\nonumber \]
We have
\[ L_r(x^r) = [x^rF(r)]_r \nonumber \]
or
\[ L(x^r \ln r) = F(r) x^r \ln x + x^r F_r(r) = (r + 3)^2 x^r \ln x +2x^r (r + 3). \nonumber \]
Now plug in \(r = -3\) to get
\[ L(x ^{-3} \ln x) = 0 .\nonumber \]
Hence
\[ y_2 = x^{ -3} \ln x. \nonumber \]
The general solution is
\[ y = c_1x ^{-3} + c_2x^{ -3} \ln x. \nonumber \]
Solve
\[ x^2y'' + 5xy' + 8y = 0 . \nonumber \]
Solution
Let
\[ y = x^r . \nonumber \]
We have
\[ F(r) = r^2 + 4r + 8 . \nonumber \]
which has complex roots
\[ r = 2 + 4i \;\;\; \text{ and } \;\;\; r = 2 - 4i \nonumber \]
We get the solutions
\[ y_1 = x^{2 + 4i} \;\;\; \text{and} \;\;\; y_2 = x^{2 - 4i} \nonumber \]
As with constant coefficients, we would like to express the solution without complex numbers. We have
\[ x ^{-2 + 2i} = e^{(-2 + 2i)\ln x} = x^{-2}e^{2\ln x i} = x^{-2}[\cos(2\ln x) + i \;\sin(2\ln x)]. \nonumber \]
Similarly
\[ x^{ -2 - 2i} = e^{(-2 - 2i)\ln x} = x^{ -2}e^{-2\ln x i} = x^{ -2}[\cos(2\ln x) - i\; \sin(2\ln x)] . \nonumber \]
By playing with constants we get the two solutions
\[ y_1 = x^{ -2} \cos(2\ln x) \;\;\; \text{and} \;\;\; y_2 = x ^{-2} \sin(2\ln x) \nonumber \]
The general solution is
\[ y = c_1x ^{-2} \cos(2\ln x) + c_2x ^{-2} \sin(2\ln x) = x^{ -2}[c_1\cos(2\ln x) + c_2\sin(2\ln x)] \nonumber \]
In summary, we have the following theorem.
Let
\[ x^2y'' + axy' + by = 0 \nonumber \]
and let
\[ F(r) = r^2 + (a - 1)r + b \nonumber \]
have roots \( r_1 \) and \( r_2 \).
- Case 1: If \( r_1 \) and \( r_2 \) are real and distinct, then the general solution is
\[ y = c_1x^{r_1} + c_2x^{r_2}. \nonumber \]
- Case 2: If \(r_1 = r_2 = r \) then the general solution is
\[ y = c_1x^r + c_2x^r \ln x. \nonumber \]
- Case 3: If \( r_1 = l + mi \) and \( r_2 = l - mi \) then the general solution is
\[ y = x^l (c_1 \cos(m\; \ln x) + c_2 \sin(m\; \ln x)). \nonumber \]
Contributors and Attributions
- Larry Green (Lake Tahoe Community College)
Integrated by Justin Marshall.