22 Jun 2025
Cauchy-Euler and Simultaneous Differential Equations
Power-law solutions of the Cauchy-Euler equation and normal combinations for coupled first- and second-order systems.
Cauchy-Euler equation
A second-order Cauchy-Euler equation has the form
\[ax^2y''+bxy'+cy=0, \qquad x>0.\]Every derivative of a power reduces its exponent by one, while the prefactor restores it. With $y=x^m$,
\[y'=mx^{m-1}, \qquad y''=m(m-1)x^{m-2}.\]Substitution gives
\[x^m\left[am(m-1)+bm+c\right]=0.\]Thus the indicial equation is
\[\boxed{am(m-1)+bm+c=0}.\]For
\[x^2y''-3xy'+4y=0,\]the indicial polynomial is
\[m(m-1)-3m+4=(m-2)^2.\]The repeated-root solution is
\[\boxed{y=x^2(C_1+C_2\ln x)}.\]The logarithm follows directly from the change of variable $t=\ln x$. Since
\[\frac{dy}{dx}=\frac1x\frac{dy}{dt},\]another differentiation gives
\[\frac{d^2y}{dx^2} =\frac1{x^2}\left(\frac{d^2y}{dt^2}-\frac{dy}{dt}\right).\]The original equation becomes
\[\frac{d^2y}{dt^2}-4\frac{dy}{dt}+4y=0,\]whose repeated root is $2$. Hence $y=(C_1+C_2t)e^{2t}=x^2(C_1+C_2\ln x)$.
Simultaneous first-order equations
Consider
\[x'=3x+y, \qquad y'=x+3y.\]Add and subtract the equations. With
\[u=x+y, \qquad v=x-y,\]we obtain
\[u'=4u, \qquad v'=2v.\]Therefore $u=Ae^{4t}$ and $v=Be^{2t}$. Solving $u=x+y$, $v=x-y$ for the original variables gives
\[\boxed{ x=\frac12(Ae^{4t}+Be^{2t}), \qquad y=\frac12(Ae^{4t}-Be^{2t})}.\]Simultaneous second-order equations
For
\[x''+2x-y=0, \qquad y''+2y-x=0,\]the same combinations decouple the system. Addition gives
\[u''+u=0,\]while subtraction gives
\[v''+3v=0.\]Thus
\[u=A\cos t+B\sin t, \qquad v=C\cos(\sqrt3t)+D\sin(\sqrt3t),\]and
\[\boxed{x=\frac{u+v}{2},\qquad y=\frac{u-v}{2}}.\]
Direct substitution into both coupled systems is carried out in the Unit II Maxima worksheet.
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