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.

bsc semester-ii mathematical-physics cauchy-euler-equation simultaneous-equations

Cauchy-Euler equation

A second-order Cauchy-Euler equation has the form

\[ax^2y^{\prime\prime}+bxy^{\prime}+cy=0, \qquad x>0.\]

Every derivative of a power reduces its exponent by one, while the prefactor restores it. With $y=x^m$,

\[y^{\prime}=mx^{m-1}, \qquad y^{\prime\prime}=m(m-1)x^{m-2}.\]

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

\[x^2y^{\prime\prime}-3xy^{\prime}+4y=0,\]

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

\[x^{\prime}=3x+y, \qquad y^{\prime}=x+3y.\]

Add and subtract the equations. With

\[u=x+y, \qquad v=x-y,\]

we obtain

\[u^{\prime}=4u, \qquad v^{\prime}=2v.\]

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

\[x^{\prime\prime}+2x-y=0, \qquad y^{\prime\prime}+2y-x=0,\]

the same combinations decouple the system. Addition gives

\[u^{\prime\prime}+u=0,\]

while subtraction gives

\[v^{\prime\prime}+3v=0.\]

Thus

\[u=A\cos t+B\sin t, \qquad v=C\cos(\sqrt3t)+D\sin(\sqrt3t),\]

and

\[\boxed{x=\frac{u+v}{2},\qquad y=\frac{u-v}{2}}.\]

The combinations \(u=x+y\) and \(v=x-y\) isolate the in-phase and out-of-phase motions. Their equations contain no remaining coupling.

Direct substitution into both coupled systems is carried out in the Unit II Maxima worksheet.

Solved Problems

1. A Cauchy-Euler initial-value problem

Solve, for $x>0$,

\[x^2y^{\prime\prime}+xy^{\prime}-4y=0, \qquad y(1)=5, \qquad y^{\prime}(1)=6.\]

With $y=x^m$, the indicial equation is

\[m(m-1)+m-4=m^2-4=0.\]

The roots are $m=2,-2$, so

\[y=C_1x^2+C_2x^{-2}.\]

At $x=1$,

\[C_1+C_2=5, \qquad 2C_1-2C_2=6.\]

Hence $C_1=4$ and $C_2=1$, giving

\[\boxed{y=4x^2+x^{-2}}.\]

Substitution gives $x^2y^{\prime\prime}+xy^{\prime}-4y=0$, and both initial conditions are recovered. The restriction $x>0$ is consistent with the logarithmic transformation used for the general theory.

2. A first-order system solved by elimination

Solve

\[x^{\prime}=x+y, \qquad y^{\prime}=4x+y, \qquad x(0)=1, \qquad y(0)=0.\]

From the first equation, $y=x^{\prime}-x$. Differentiating and using the second equation gives

\[x^{\prime\prime}-x^{\prime}=4x+(x^{\prime}-x),\]

or

\[x^{\prime\prime}-2x^{\prime}-3x=0.\]

Its roots are $3$ and $-1$, so

\[x=Ae^{3t}+Be^{-t}, \qquad y=x^{\prime}-x=2Ae^{3t}-2Be^{-t}.\]

The initial data give $A+B=1$ and $A-B=0$. Therefore

\[\boxed{ x=\frac{e^{3t}+e^{-t}}2, \qquad y=e^{3t}-e^{-t}}.\]

Direct substitution makes both first-order residuals zero. The growing $e^{3t}$ normal component eventually dominates the decaying $e^{-t}$ component.

3. Two second-order normal combinations

Solve

\[x^{\prime\prime}+3x-y=0, \qquad y^{\prime\prime}+3y-x=0,\]

subject to $x(0)=1$, $y(0)=0$, and $x^{\prime}(0)=y^{\prime}(0)=0$.

Set $u=x+y$ and $v=x-y$. Adding and subtracting the equations gives

\[u^{\prime\prime}+2u=0, \qquad v^{\prime\prime}+4v=0.\]

The initial data become $u(0)=v(0)=1$ and $u^{\prime}(0)=v^{\prime}(0)=0$, hence

\[u=\cos(\sqrt2t), \qquad v=\cos2t.\]

Recovering the original variables,

\[\boxed{ x=\frac{\cos(\sqrt2t)+\cos2t}{2}, \qquad y=\frac{\cos(\sqrt2t)-\cos2t}{2}}.\]

Both coupled equations and all four initial data are satisfied. If $t$ is measured in seconds, the two normal angular frequencies are $\sqrt2\,\mathrm{rad\,s^{-1}}$ and $2\,\mathrm{rad\,s^{-1}}$.

Descriptive Questions

1. Derive the indicial equation for a second-order Cauchy-Euler equation and relate it to the substitution $t=\ln x$.
2. Explain the origin of the $x^m\ln x$ term when the indicial equation has a repeated root.
3. Show how elimination converts two simultaneous first-order equations into one second-order equation.
4. Explain why normal combinations decouple a symmetric pair of simultaneous second-order equations.

Numerical Problems

1. For $x^2y^{\prime\prime}+xy^{\prime}+9y=0$, find the logarithmic phase period and the factor by which $x$ must be multiplied to repeat the phase. Final answer: $m=\pm3i$, $\Delta(\ln x)=2\pi/3$, and $x_2/x_1=e^{2\pi/3}=8.1205$.
2. Find the normal growth rates of $x^{\prime}=2x+3y$, $y^{\prime}=3x+2y$. Final answer: the rates are $5$ for $x+y$ and $-1$ for $x-y$.
3. Find the normal angular frequencies of $x^{\prime\prime}+5x-2y=0$, $y^{\prime\prime}+5y-2x=0$. Final answer: $\omega_{+}=\sqrt3$ and $\omega_{-}=\sqrt7$ in inverse-time units.
4. Solve $x^2y^{\prime\prime}-5xy^{\prime}+9y=0$ with $y(1)=2$, $y^{\prime}(1)=7$, and evaluate $y(2)$. Final answer: $y=x^3(2+\ln x)$; $y(2)=8(2+\ln2)=21.5452$.

The three solution residuals and all numerical values are verified in the Unit II problem-check worksheet.

References

1. Cauchy-Euler equation — Wikipedia
2. William E. Boyce, Richard C. DiPrima, and Douglas B. Meade, Elementary Differential Equations and Boundary Value Problems, 11th ed., chapters “Higher-Order Linear Equations” and “Systems of First-Order Linear Equations,” Wiley.
3. Erwin Kreyszig, Advanced Engineering Mathematics, 10th ed., chapters “Second-Order Linear ODEs” and “Systems of ODEs,” Wiley.
4. Mary L. Boas, Mathematical Methods in the Physical Sciences, 3rd ed., Chapter 8, “Ordinary Differential Equations,” Wiley.