- Cauchy-Euler equation
In
mathematics , a Cauchy-Euler equation (also Euler-Cauchy equation) is a linear homogeneousordinary differential equation withvariable coefficient s. They are sometimes known as equi-dimensional equations. Because of its simple structure the equation can be replaced with an equivalent equation withconstant coefficient s which can then be solved explicitly.The equation
Let y^{(n)}(x) be the "n"th derivative of the unknown function y(x). Then a Cauchy-Euler equation of order "n" has the form
:x^n y^{(n)}(x) + a_{n-1} x^{n-1} y^{(n-1)}(x) + cdots + a_0 y(x) = 0.
The substitution scriptstyle x = e^u reduces this equation to a linear differential equation with constant coefficients.
Second order
The Euler-Cauchy equation crops up in a number of engineering applications. It is given by the equation:
:x^2frac{d^2y}{dx^2} + axfrac{dy}{dx} + by = 0 ,
We assume a trial solution given by
:y = x^m. ,
Differentiating, we have:
:frac{dy}{dx} = mx^{m-1} ,
and
:frac{d^2y}{dx^2} = m(m-1)x^{m-2}. ,
Substituting into the original equation, we have:
:x^2( m(m-1)x^{m-2} ) + ax( mx^{m-1} ) + b( x^m ) = 0 ,
Or rearranging gives:
:m^2 + (a-1)m + b = 0. ,
We then can solve for "m". There are three particular cases of interest:
* Case #1: Two distinct roots, "m"1 and "m"2
* Case #2: One real repeated root, "m"
* Case #3: Complex roots, α ± "i"βIn case #1, the solution is given by::y = c_{1}x^{m_{1 + c_{2}x^{m_{2 ,
In case #2, the solution is given by::y = c_{1}x^{m}ln(x) + c_{2}x^{m} ,To get to this solution, the method of reduction of order must be applied after having found one solution "y" = "x""m".
In case #3, the solution is given by:
:y = c_{1}x^{alpha}cos(eta ln(x)) + c_{2}x^{alpha}sin(eta ln(x)) ,
This equation also can be solved with x = e^t transformation.
This particular case is of no great practical importance and hence this has been left as a challenge for the reader.
Example
Given : x^2u"-3xu'+3u=0,we substitute the simple solution "x"α:: x^2(alpha(alpha-1)x^{alpha-2})-3x(alpha(x^{alpha-1}))+3(x^alpha)=alpha(alpha-1)x^alpha-3alpha x^alpha+3x^alpha,: alpha(alpha-1)-3alpha+3)x^alpha.,For this to indeed be a solution, either "x"=0 giving the trivial solution, or the coefficient of "x"α is zero, so solving that quadratic, we get α=1,3. So, the general solution is: u=Ax+Bx^3.,
Difference equation analogue
There is a
difference equation analogue to the Cauchy–Euler equation. For a fixed m > 0, define the sequence f_m(n) as: f_m(n) := n(n+1)cdots (n+m-1).
Applying the difference operator to f_m, we find that
: Df_m(n) = f_{m+1}(n) - f_m(n) = m(n+1)(n+2) cdots (n+m-1) = frac{m}{n} f_m(n).
If we do this "k" times, we will find that
: f_m^{(k)}(n) = frac{m(m-1)cdots(m-k+1)}{n(n+1)cdots(n+k-1)} f_m(n) = m(m-1)cdots(m-k+1) frac{f_m(n)}{f_k(n)},
where the superscript ("k") denotes applying the difference operator "k" times. Comparing this to the fact that the "k"-th derivative of x^m equals m(m-1)cdots(m-k+1)frac{x^m}{x^k} suggests that we can solve the "N"-th order difference equation
: f_N(n) y^{(N)}(n) + a_{N-1} f_{N-1}(n) y^{(N-1)}(n) + cdots + a_0 y(n) = 0,
in a similar manner to the differential equation case. Indeed, substituting the trial solution
: y(n) = f_m(n)
brings us to the same situation as the differential equation case,
: m(m-1)cdots(m-N+1) + a_{N-1} m(m-1) cdots (m-N+2) + cdots + a_1 m + a_0 = 0.
One may now proceed as in the differential equation case, since the general solution of an "N"-th order linear difference equation is also the linear combination of "N" linearly independent solutions. Applying reduction of order in case of a multiple root m_1 will yield expressions involving a discrete version of ln,
: varphi(n) = sum_{k=1}^n frac{1}{k - m_1}.
(Compare with: ln (x - m_1) = int_{1+m_1}^x frac{1}{t - m_1} , dt.)
In cases where fractions become involved, one may use
: f_m(n) := frac{Gamma(n+m)}{Gamma(n)}
instead (or simply use it in all cases), which coincides with the definition before for integer "m".
ee also
*
Hypergeometric differential equation
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