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5.7: Cauchy-Euler Equations

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    108310
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    A special class of linear differential equations that is of interest are Cauchy-Euler equations, defined as follows.

    Definition 5.7.1 Cauchy-Euler Equations

    A second order Cauchy-Euler equation is an equation that can be written in the form

    \[\label{eq:7.4.6} ax^2y''+bxy'+cy=0,\]

    where \(a,b\), and \(c\) are real constants and \(a\ne0\).

    Theorem 5.1.1 implies that Equation \ref{eq:7.4.6} has solutions defined on \((0,\infty)\) and \((-\infty,0)\), since Equation \ref{eq:7.4.6} can be rewritten as

    \[ay''+{b\over x}y'+{c\over x^2}y=0. \nonumber\]

    For convenience we’ll restrict our attention to the interval \((0,\infty)\). (Exercise 7.4.40 deals with solutions of Equation \ref{eq:7.4.6} on \((-\infty,0)\).)

    The key to finding solutions on \((0,\infty)\) is that if \(x>0\) then we can guess that \(y=x^r\) is defined as a real-valued function on \((0,\infty)\) for all values of \(r\), and it may lead to a solution. Substituting \(y=x^r\) into Equation \ref{eq:7.4.6} produces

    \[\label{eq:7.4.7} \begin{array}{lcl} ax^2(x^r)''+bx(x^r)'+cx^r&= ax^2r(r-1)x^{r-2}+bxrx^{r-1}+cx^r\\ &= [ar(r-1)+br+c]x^r. \end{array}\]

    The equation \[\label{eq:7.4.8}ar(r-1)+br+c=ar^2+(b-a)r+c=0\]

    is called the characteristic equation of Equation \ref{eq:7.4.6}. From Equation \ref{eq:7.4.7} we can see that \(y=x^r\) is a solution of Equation \ref{eq:7.4.6} on \((0,\infty)\) if and only if \(ar(r-1)+br+c=0\).

    The roots of the characteristic equation are given by the quadratic formula

    \[\label{eq:7.4.9} r={-(b-a)\pm\sqrt{(b-a)^2-4ac}\over2a}.\]

    Three Cases

    As you'll recall from algebra, there are three possible types of solutions to \ref{eq:7.4.8} and we will exam each case separately

    • Case 1: \((b-a)^2-4ac>0\), so the characteristic equation has two distinct real roots.
    • Case 2: \((b-a)^2-4ac=0\), so the characteristic equation has a repeated real root.
    • Case 3: \((b-a)^2-4ac<0\), so the characteristic equation has complex roots.

    Case 1: Distinct Real Roots

    Let \(r_1\) and \(r_2\) be distinct real roots of \ref{eq:7.4.8}.

    Therefore, \(y_1=x^{r_1}\) and \(y_2=x^{r_2}\) form a fundamental set of solutions of Equation \ref{eq:7.4.6} on \((0,\infty)\), since \(y_2/y_1=x^{r_2-r_1}\) is nonconstant. In this case

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

    is the general solution of Equation \ref{eq:7.4.6} on \((0,\infty)\).

    Note: we could also use the Wronskian to show independence:

    \[W(x)=\left| \begin{array}{rr} x^{r_1} & x^{r_2} \\ r_1x^{r_1-1} & r_2x^{r_2-1} \end{array} \right|=(r_2-r_1)x^{(r_1+r_2-1)}\not\equiv0\nonumber\] for all x on \((0,\infty)\) since \(r_1\ne r_2\).

    Example 5.7.1

    Find the general solution of

    \[\label{eq:7.4.10} x^2y''-xy'-8y=0\]

    on \((0,\infty)\).

    Solution

    The characteristic equation of \ref{eq:7.4.10} is

    \[r(r-1)-r-8=(r-4)(r+2)=0. \nonumber\]

    Therefore \(y_1=x^4\) and \(y_2=x^{-2}\) are solutions of Equation \ref{eq:7.4.10} on \((0,\infty)\), and its general solution on \((0,\infty)\) is

    \[y=c_1x^4+{c_2\over x^2}.\nonumber\]

    Example 5.7.2

    Find the general solution of

    \[\label{eq:7.4.11} 6x^2y''+5xy'-y=0\]

    on \((0,\infty)\).

    Solution

    The characteristic equation of \ref{eq:7.4.11} is

    \[6r(r-1)+5r-1=(2r-1)(3r+1)=0.\nonumber\]

    Therefore the general solution of Equation \ref{eq:7.4.11} on \((0,\infty)\) is

    \[y=c_1x^{1/2}+c_2x^{-1/3}. \nonumber \]

    Case 2: A Repeated Real Root

    Let \(r\) be a repeated real root of \ref{eq:7.4.8}. As such, we only get one solution, \(y_1=x^{r}\). But we know there must be two independent solutions to a second order homogeneous differential equation. So, how can we find a second independent solution if we only know one? Recall the formula for a second solution that we derived in section 5.2,

    \[y_2=y_1\int{e^{-\int{p(x)dx}}\over y_1^2}dx\nonumber\]

    If we divide \ref{eq:7.4.7} by \(a\) we get \[y''+{b\over ax}y'+{c\over ax^2}y=0\nonumber\]

    which gives is \(p(x)={b\over ax}\), which gives us \[y_2=x^{r}\int{e^{-\int{b\over ax}dx}\over x^{2r}}dx\nonumber\]

    \[=x^{r}\int{x^{-{b\over a}}\over x^{2r}}dx\nonumber\]

    Since we have a repeated root we know \((b-a)^2-4ac=0\) in \ref{eq:7.4.9} which gives us \(r={-(b-a)\over 2a}\) or

    \[{-b\over a}=2r-1\nonumber\]

    Using this we now get \[y_2=x^{r}\int{x^{2r-1}\over x^{2r}}dx\nonumber\]

    \[=x^{r}\int {1\over x}dx\nonumber\]

    \[=x^r \ln x\nonumber\]

    Since we already know that the formula gives us a second linearly independent solution, the general solution is \[y=c_1x^{r}+c_2x^r\ln x\] on \((0,\infty)\).

    Example \(\PageIndex{3}\)

    Find the general solution of

    \[\label{eq:7.4.12} x^2y''-5xy'+9y=0\]

    on \((0,\infty)\).

    Solution

    The characteristic equation of \ref{eq:7.4.12} is

    \[r(r-1)-5r+9=(r-3)^2=0.\nonumber\]

    Therefore the general solution of Equation \ref{eq:7.4.12} on \((0,\infty)\) is

    \[y=x^3(c_1+c_2 \ln x).\nonumber\]

    Example \(\PageIndex{4}\)

    a. Find the general solution of

    \[\label{eq:7.4.13} x^2y''+3xy'+y=0\]

    on \((0,\infty)\).

    b. Solve the initial value problem

    \[\label{eq:7.4.14} x^2y''+3xy'+y=0. \quad y(1)=0, y'(1)=1\]

    on \((0,\infty)\).

    Solution a

    The characteristic equation of \ref{eq:7.4.13} is

    \[r(r-1)+3r+1=(r+1)^2=0.\nonumber\]

    Therefore the general solution of Equation \ref{eq:7.4.13} on \((0,\infty)\) is

    \[y=x^{-1}(c_1+c_2 \ln x).\nonumber\]

    Solution b

    At \(x=1\) we know the values of \(y\) and \(y^{\prime}\). Using the general solution of \ref{eq:7.4.13}, we first have that

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

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

    Now, using the second condition and

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

    we have

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

    Therefore, the solution of \ref{eq:7.4.14} on \((0,\infty)\) is \[y=x^{-1}\ln x\nonumber\].

    Case 3: Complex Conjugate Roots

    Now suppose \ref{eq:7.4.8} has complex conjugate roots \(r=\lambda \pm \omega i\). This is similar to case one in that we do have distinct roots and we could say \(y_1=x^{(\lambda + \omega i)}\) and \(y_2=x^{(\lambda - \omega i)}\). The problem with this is that it's not very useful in real life applications to have functions of complex numbers. So, we want to see if we can somehow write both solutions as functions of real numbers, and as we did earlier in the chapter we will use Euler's Identity.

    We will now rewrite \(y_1\) and \(y_2\) as follows:

    \[y_1=x^{(\lambda + \omega i)}=e^{\ln x^{\lambda+\omega i}}=e^{(\lambda+\omega i)\ln x}=e^{\lambda ln x}e^{i\omega \ln x}=e^{\lambda ln x}(\cos (\omega \ln x)+i\sin (\omega \ln x))=e^{\lambda ln x}\cos (\omega \ln x)+ie^{\lambda ln x}\sin (\omega \ln x)\nonumber\]

    \[y_2=x^{(\lambda - \omega i)}=e^{\ln x^{\lambda-\omega i}}=e^{(\lambda-\omega i)\ln x}=e^{\lambda ln x}e^{-i\omega \ln x}=e^{\lambda ln x}(\cos (-\omega \ln x)+i\sin (-\omega \ln x))=e^{\lambda ln x}\cos (\omega \ln x)-ie^{\lambda ln x}\sin (\omega \ln x)\nonumber\]

    The latter is using the fact the \(\cos\) is an even function and \(\sin\) is an odd function.

    Note: the \(i\) is no longer inside a function.

    Recall that if \(y_1\) and \(y_2\) are solutions of a homogeneous equation, then any linear combination is also a solution. So, if we add \(y_1\) and \(y_2\) we get

    \[y_3=2e^{\lambda ln x}\cos (\omega \ln x)=2e^{ln x^{\lambda}}\cos (\omega \ln x)=2x^{\lambda}\cos (\omega \ln x)\nonumber\]

    and if we subtract \(y_1\) and \(y_2\) we get

    \[y_4=2ie^{\lambda ln x}\sin(\omega \ln x)=2ie^{ln x^{\lambda}}\sin(\omega \ln x)=2ix^{\lambda}\sin (\omega \ln x)\nonumber\]

    Now, both 2 and 2i are constants that can be absorbed into the constants \(c_1\) and \(c_2\) of our general solution and since \(y_3\) and \(y_4\) are not constant multiples of each other, they are independent. Thus, our general solution is \[y=c_1x^{\lambda}\cos (\omega \ln x)+c_2x^{\lambda}\sin (\omega \ln x)\]

    Example 5.7.5

    Find the general solution of

    \[\label{eq:7.4.15} x^2y''+3xy'+2y=0\]

    on \((0,\infty)\).

    Solution

    The characteristic equation of \ref{eq:7.4.15} is

    \[r(r-1)+3r+2=(r+1)^2+1=0.\nonumber\]

    The roots of the characteristic equation are \(r=-1 \pm i\) and the general solution of Equation \ref{eq:7.4.15} on \((0,\infty)\) is

    \[y={1\over x}\left[c_1\cos(\ln x)+c_2\sin(\ln x)\right].\nonumber\]

    Example \(\PageIndex{6}\)

    Find the general solution of

    \[\label{eq:7.4.16} x^2y''-xy'+5y=0\]

    on \((0,\infty)\).

    Solution

    The characteristic equation of \ref{eq:7.4.16} is

    \[r(r-1)-r+5=(r-1)^2+4=0.\nonumber\]

    The roots of the characteristic equation are \(r=1 \pm 2i\) and the general solution of Equation \ref{eq:7.4.16} on \((0,\infty)\) is

    \[y=x\left[c_1\cos(2\ln x)+c_2\sin(2\ln x)\right].\nonumber\]

    Nonhomogeneous Cauchy-Euler Equations

    We can solve nonhomogeneous Cauchy-Euler equations using the Method of Variation of Parameters. We will demonstrate this with a couple examples.

    Example \(\PageIndex{7}\)

    Find the general solution of \[\label{eq:7.4.17} x^{2} y^{\prime \prime}-x y^{\prime}-3 y=2 x^{2}\] on \((0,\infty)\).

    Solution

    First we find the solution of the homogeneous equation

    \[\label{eq:7.4.18} x^{2} y^{\prime \prime}-x y^{\prime}-3 y=0.\]

    The characteristic equation of \ref{eq:7.4.18} is \(r^{2}-2 r-3=0 .\) So, the roots are \(r=-1,3\) and the solution of \ref{eq:7.4.18} is

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

    We set

    \[y_p=u_1x^{-1}+u_2x^3, \nonumber \]

    where

    \[\begin{aligned} x^{-1}u_1'+x^3u_2'&=0\\ \phantom{x}-x^{-2}u_1'+3x^2u_2'&={2x^2\over x^2}=2.\end{aligned}\]

    Now, letting \[W=\left| \begin{array}{cc} x^{-1} & x^3 \\ -x^{-2} & 3x^2 \end{array} \right|=4x,\quad W_1=\left| \begin{array}{cc} 0 & x^3 \\ 2 & 3x^2 \end{array} \right|=-2x^3, \quad W_2=\left| \begin{array}{cc} x^{-1} & 0 \\ -x^{-2} & 2 \end{array} \right|=2x^{-1},\nonumber \]

    So, \(u_1'={W_1\over W}=-{x^2\over 2}\) and \(u_2'={W_2\over W}={x^{-2}\over 2}\). Integrating and taking the constants of integration to be zero yields

    \[u_1=-{x^3\over 6} \quad \text{and} \quad u_2=-{x^{-1}\over 2}. \nonumber \]

    Therefore

    \[y_p=u_1x^{-1}+u_2x^3 =-{x^3\over 6}x^{-1}-{x^{-1}\over 2}x^3=-{2\over 3}x^2, \nonumber \]

    so the general solution of Equation \ref{eq:7.4.17} is

    \[y=y_p+c_1x^{-1}+c_2x^3=-{2\over 3}x^2+c_1x^{-1}+c_2x^3\nonumber\] on \((0,\infty)\).

    Example \(\PageIndex{8}\)

    Find the general solution of \[\label{eq:7.4.19} x^{2} y^{\prime \prime}-x y^{\prime}-3 y=2 x^3\] on \((0,\infty)\).

    Solution

    \ref{eq:7.4.19} has the same homogenous solution as \ref{eq:7.4.17}, so

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

    We set

    \[y_p=u_1x^{-1}+u_2x^3, \nonumber \]

    where

    \[\begin{aligned} x^{-1}u_1'+x^3u_2'&=0\\ \phantom{x}-x^{-2}u_1'+3x^2u_2'&={2x^2\over x^2}=2x.\end{aligned}\]

    Now, letting \[W=\left| \begin{array}{cc} x^{-1} & x^3 \\ -x^{-2} & 3x^2 \end{array} \right|=4x,\quad W_1=\left| \begin{array}{cc} 0 & x^3 \\ 2x & 3x^2 \end{array} \right|=-2x^4, \quad W_2=\left| \begin{array}{cc} x^{-1} & 0 \\ -x^{-2} & 2x \end{array} \right|=2,\nonumber \]

    So, \(u_1'={W_1\over W}=-{x^3\over 2}\) and \(u_2'={W_2\over W}={2\over 4x}\). Integrating and taking the constants of integration to be zero yields

    \[u_1=-{x^4\over 8} \quad \text{and} \quad u_2={1\over 2}\ln x. \nonumber \]

    Therefore

    \[y_p=u_1x^{-1}+u_2x^3 =-{x^4\over 8}x^{-1}+({1\over 2}\ln x)x^3=-{x^3\over 8}+{x^3\over 2}\ln x, \nonumber \]

    However, since \(c_2\) is an arbitrary constant, so is \(c_2-{1\over 8}\); therefore, we improve the appearance of this result by renaming the constant and writing the general solution of Equation \ref{eq:7.4.19} as

    \[y=y_p+c_1x^{-1}+c_2x^3={x^3\over 2}\ln x+c_1x^{-1}+c_2x^3\nonumber\] on \((0,\infty)\).


    This page titled 5.7: Cauchy-Euler Equations is shared under a CC BY-NC-SA 1.3 license and was authored, remixed, and/or curated by Russell Herman via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.