Exercises — Cauchy-Euler (Equidimensional) equation

The signature is: power of $x$ = order of derivative it multiplies. Look at each term.

• (a) $x^2{y}^{\prime \prime }$ (power 2, order 2 ✓), $5x{y}^{\prime }$ (power 1, order 1 ✓), $3y$ (power 0, order 0 ✓). Yes. Here $a=1,b=5,c=3$, so indicial: $m(m-1)+5m+3=0⟹m^2+4m+3=0.$
• (b) The middle term is $5y^{\prime }$ — power of $x$ is $0$ but the derivative order is $1$. Mismatch. Not Cauchy-Euler.
• (c) $3x^2{y}^{\prime \prime }$ (✓), $-x{y}^{\prime }$ (✓), $y$ (✓). Yes. $a=3,b=-1,c=1$: $3m(m-1)-m+1=0⟹3m^2-4m+1=0.$

Here $a=1,b=-7,c=16$. Indicial: $m(m-1)-7m+16=0⟹m^2-8m+16=0.$ The discriminant is $b^{\prime 2}-4a^{\prime }{c}^{\prime }=(-8{)}^2-4(1)(16)=64-64=0$. A zero discriminant means one repeated root — Case 2, so a $\ln x$ is coming.

$a=1,b=-2,c=-4$. Indicial: $m(m-1)-2m-4=m^2-3m-4=0$. Factor: $(m-4)(m+1)=0\Rightarrow m=4,-1$. Distinct real roots → two independent powers: $\boxed{y=C_1{x}^4+C_2{x}^{-1}}$

$a=1,b=-3,c=4$. Indicial: $m(m-1)-3m+4=m^2-4m+4=(m-2{)}^2=0\Rightarrow m=2$ (double). One power $x^2$; the partner picks up a $\ln x$ (from the $t{e}^{mt}$ solution in the $t=\ln x$ world): $\boxed{y=(C_1+C_2\ln x)x^2}$

$a=1,b=1,c=1$. Indicial: $m(m-1)+m+1=m^2+1=0\Rightarrow m=\pm i$. So $\alpha =0,\beta =1$. Since $x^{\pm i}=e^{\pm i\ln x}$, Euler's Formula converts to real trig: $\boxed{y=C_1\cos (\ln x)+C_2\sin (\ln x)}$

$a=2,b=3,c=-1$. Indicial: $2m(m-1)+3m-1=2m^2+m-1=0$. Quadratic formula: $m=\frac{-1\pm \sqrt{1+8}}{4}=\frac{-1\pm 3}{4}$, giving $m=\frac{1}{2}$ and $m=-1$. $\boxed{y=C_1{x}^{1/2}+C_2{x}^{-1}}$

$a=1,b=5,c=4$. Indicial: $m(m-1)+5m+4=m^2+4m+4=(m+2{)}^2=0\Rightarrow m=-2$ (repeated). General solution: $y=(C_1+C_2\ln x)x^{-2}$. Now apply the conditions. At $x=1$, $\ln 1=0$, so $y(1)=C_1=3$. Differentiate with the product rule: $y^{\prime }=C_2\cdot \frac{1}{x}\cdot x^{-2}+(C_1+C_2\ln x)\cdot (-2)x^{-3}=[C_2-2(C_1+C_2\ln x)]x^{-3}.$ At $x=1$: $y^{\prime }(1)=C_2-2C_1=C_2-6=-11\Rightarrow C_2=-5$. $\boxed{y=(3-5\ln x)x^{-2}}$

$a=1,b=1,c=4$. Indicial: $m(m-1)+m+4=m^2+4=0\Rightarrow m=\pm 2i$. So $\alpha =0,\beta =2$. $\boxed{y=C_1\cos (2\ln x)+C_2\sin (2\ln x)}$ Behaviour. Because $\alpha =0$ there is no growing/decaying power factor $x^{\alpha }$ out front — the amplitude stays constant. The argument is $2\ln x$. As $x\to \infty$, $\ln x\to \infty$ slowly, so the curve keeps oscillating but the oscillations get stretched out (each cycle spans a wider range of $x$). As $x\to 0^{+}$, $\ln x\to -\infty$, so the curve oscillates infinitely many times, packing waves ever tighter — it never settles to a limit. See the chalkboard sketch.

$a=1,b=-5,c=8$. Indicial: $m(m-1)-5m+8=m^2-6m+8=(m-2)(m-4)=0\Rightarrow m=2,4$. On $x>0$ we would write $y=C_1{x}^2+C_2{x}^4$. But here $x<0$. A power like $x^{1/2}$ or $\ln x$ would be undefined for negative $x$, so we protect ourselves with $|x|$: solutions are built from $|x{|}^m$ (and $\ln |x|$ when needed). Here the exponents $2,4$ are even integers, so $x^2=|x{|}^2$ and $x^4=|x{|}^4$ anyway — but the safe general form is: $\boxed{y=C_1|x{|}^2+C_2|x{|}^4=C_1{x}^2+C_2{x}^4}$ The habit of writing $|x{|}^m$ matters the moment an exponent is fractional or a $\ln$ appears.

Substitute $t=\ln x$. From the parent note, $x{y}^{\prime }=\dot{y}$ and $x^2{y}^{\prime \prime }=\ddot{y}-\dot{y}$ (dots are $d/dt$). Here $a=1,b=-1,c=1$, so the transformed constant-coefficient ODE is $\ddot{y}+(b-a)\dot{y}+cy=\ddot{y}-2\dot{y}+y=t.$ (Note the right side $\ln x$ became $t$.) Homogeneous part: characteristic $m^2-2m+1=(m-1{)}^2=0\Rightarrow m=1$ (repeated), so $y_h=(C_1+C_{2}t)e^t=(C_1+C_2\ln x)x$. Particular part by undetermined coefficients: try $y_p=At+B$ (a first-degree polynomial matches the right side $t$; it does not clash with $e^t$). Then ${\dot{y}}_p=A,{\ddot{y}}_p=0$: $0-2A+(At+B)=t\Rightarrow At+(B-2A)=t.$ Match: $A=1$ and $B-2A=0\Rightarrow B=2$. So $y_p=t+2=\ln x+2$. $\boxed{y=(C_1+C_2\ln x)x+\ln x+2}$

Homogeneous: $a=1,b=-1,c=-3$. Indicial: $m(m-1)-m-3=m^2-2m-3=(m-3)(m+1)=0\Rightarrow m=3,-1$. $y_h=C_1{x}^3+C_2{x}^{-1}.$ Particular: the right side is $x^2$. Since $2$ is not a root of the indicial equation ($m=3,-1$), the power $x^2$ is not a homogeneous solution, so a plain guess $y_p=A{x}^2$ is safe (no resonance). Then $y_p^{\prime }=2Ax,y_p^{\prime \prime }=2A$. Substitute: $x^2(2A)-x(2Ax)-3(A{x}^2)=(2A-2A-3A)x^2=-3A{x}^2=x^2.$ So $-3A=1\Rightarrow A=-\frac{1}{3}$, giving $y_p=-\frac{1}{3}{x}^2$. $\boxed{y=C_1{x}^3+C_2{x}^{-1}-\frac{1}{3}{x}^2}$

Homogeneous: $a=1,b=-1,c=1$. Indicial: $m(m-1)-m+1=m^2-2m+1=(m-1{)}^2=0\Rightarrow m=1$ (repeated). $y_h=(C_1+C_2\ln x)x.$ Particular: the right side is $x^1$, and $m=1$ is a double root of the indicial equation — double resonance. In the $t$-world this is like forcing $e^t$ against $(m-1{)}^2$, which needs $t^2{e}^t$. Translating back, we try $y_p=Ax(\ln x{)}^2.$ Easiest to work in $t=\ln x$: transformed equation is $\ddot{y}-2\dot{y}+y=e^t$ (since $x=e^t$). Try $y_p=A{t}^2{e}^t$. Using $(D-1{)}^2[t^2{e}^t]=2A{e}^t$ (the two derivatives kill the polynomial down to its constant $2A$ term), we need $2A=1\Rightarrow A=\frac{1}{2}$. So $y_p=\frac{1}{2}{t}^2{e}^t=\frac{1}{2}(\ln x{)}^{2}x$. $\boxed{y=(C_1+C_2\ln x)x+\frac{1}{2}(\ln x{)}^{2}x}$

For third order, $y=x^m$ needs $x^3{y}^{\prime \prime \prime }=m(m-1)(m-2)x^m$, $x^2{y}^{\prime \prime }=m(m-1)x^m$, $x{y}^{\prime }=m{x}^m$, $y=x^m$. Indicial: $m(m-1)(m-2)+2m(m-1)-m+1=0.$ Expand: $m(m-1)(m-2)=m^3-3m^2+2m$; $2m(m-1)=2m^2-2m$. Sum with $-m+1$: $m^3-3m^2+2m+2m^2-2m-m+1=m^3-m^2-m+1=0.$ Factor by grouping: $m^2(m-1)-(m-1)=(m-1)(m^2-1)=(m-1{)}^2(m+1)=0$. Roots: $m=1$ (double), $m=-1$ (simple). Double root $\to$ pair $x^1$ and $(\ln x)x^1$; simple root $\to x^{-1}$: $\boxed{y=(C_1+C_2\ln x)x+C_3{x}^{-1}}$

The solution form $x^{\alpha }[\dots \cos (\beta \ln x)\dots ]$ tells us the roots are $m=\alpha \pm i\beta$ with $\alpha =3,\beta =1$, i.e. $m=3\pm i$. The indicial polynomial with those roots is $(m-(3+i))(m-(3-i))=(m-3{)}^2+1=m^2-6m+10.$ Now match to $a{m}^2+(b-a)m+c$ with $a=1$: we need $b-a=-6$ and $c=10$, so $b=-5$, $c=10$. $\boxed{x^2{y}^{\prime \prime }-5x{y}^{\prime }+10y=0}$ Check: indicial $m(m-1)-5m+10=m^2-6m+10=0\Rightarrow m=3\pm i$. ✓

(i) Here $a=1,b=0,c=-2$. Indicial: $m(m-1)-2=m^2-m-2=(m-2)(m+1)=0\Rightarrow m=2,-1$. $\boxed{y=C_1{x}^2+C_2{x}^{-1}}$ (ii) Write the ODE in standard form $y^{\prime \prime }+p(x)y^{\prime }+q(x)y=0$ by dividing by $x^2$: $y^{\prime \prime }+0\cdot y^{\prime }-\frac{2}{x^2}y=0.$ The coefficient $q(x)=-2/x^2$ blows up at $x=0$: the leading coefficient $x^2$ of the original equation vanishes there, so $x=0$ is a singular point. Notice the solution $C_2{x}^{-1}$ literally goes to $\pm \infty$ as $x\to 0$ — you cannot pose a well-behaved initial condition at $x=0$. Because $x\cdot p(x)=0$ and $x^2\cdot q(x)=-2$ stay finite as $x\to 0$, it is a regular singular point — exactly the setting where the Frobenius method applies. In fact Cauchy-Euler is the simplest Frobenius case: the indicial equation here is the Frobenius indicial equation, and the power solutions $x^m$ are the leading terms of Frobenius series with no correction terms needed.

Distinct real roots $:::y=C_1{x}^{m_1}+C_2{x}^{m_2}$, no logs. Repeated root $:::y=(C_1+C_2\ln x)x^m$, one log. Complex roots $\alpha \pm i\beta$ $:::y=x^{\alpha }[C_1\cos (\beta \ln x)+C_2\sin (\beta \ln x)]$. Right side $x^r$ with $r$ a simple indicial root $:::y_p$ needs an extra $\ln x$ factor. Right side $x^r$ with $r$ a double indicial root $:::y_p$ needs $(\ln x{)}^2$.