Cauchy–Euler Equation with a Resonant Power-Law Forcing: 04-BS-1, December 2014, Q2

Question 2: Cauchy–Euler Equation with a Resonant Power-Law Forcing (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A Cauchy–Euler (equidimensional) ODE $2x^2y''+xy'-3y=4x^{-1}$, $x>0$.

Find. The general solution $y(x)$.

Approach. Try $y=x^m$ for the homogeneous equation to get the characteristic (indicial) polynomial in $m$. Since the forcing $4x^{-1}$ turns out to duplicate one of the homogeneous exponents, use the resonant Cauchy–Euler trial $y_p=Ax^{-1}\ln x$ instead of the usual $Ax^{-1}$.

Homogeneous (indicial) equation. Substituting $y=x^m$ into $2x^2y''+xy'-3y=0$ gives $2m(m-1)+m-3=0\Rightarrow2m^2-m-3=0=(2m-3)(m+1)$, so $m=\tfrac32,-1$. $$y_h=C_1x^{3/2}+C_2x^{-1}.$$
Resonance check. The forcing $4x^{-1}$ matches the homogeneous mode $x^{-1}$ ($m=-1$), so the ordinary trial $y_p=Ax^{-1}$ solves to $0=4x^{-1}$ — instead use $y_p=Ax^{-1}\ln x$.
Differentiate the trial. $$y_p'=Ax^{-2}(1-\ln x),\qquad y_p''=Ax^{-3}(2\ln x-3).$$
Substitute and solve for $A$. $$2x^2y_p''+xy_p'-3y_p=Ax^{-1}\big[2(2\ln x-3)+(1-\ln x)-3\ln x\big]=Ax^{-1}(-5)=-5Ax^{-1}.$$ Setting $-5A=4$ gives $A=-\tfrac45$. $$y_p=-\tfrac45x^{-1}\ln x.$$

$$y(x)=\boxed{C_1x^{3/2}+C_2x^{-1}-\tfrac45x^{-1}\ln x}$$

Quantity	Result
Indicial roots	$m=3/2,\ -1$
Particular solution	$-\tfrac45x^{-1}\ln x$
General solution	$y(x)=C_1x^{3/2}+C_2x^{-1}-\tfrac45x^{-1}\ln x$

Concept & Theory

Topic. Cauchy–Euler (equidimensional) ODEs and resonance at a repeated/matching indicial root.

Key relations. For $ax^2y''+bxy'+cy=0$, the trial $y=x^m$ gives the indicial equation $am(m-1)+bm+c=0$; if a power-law forcing term matches one of the indicial roots exactly, the usual trial $Ax^m$ fails and must be replaced by $Ax^m\ln x$ (directly analogous to the $xe^{kx}$ fix for constant-coefficient resonance).

Why this works. A Cauchy–Euler equation transforms into a constant-coefficient ODE under $x=e^{t}$; the power-law forcing $x^{-1}$ becomes $e^{-t}$ in that variable, which is exactly a resonant exponential forcing there, so the $\ln x=t$ multiplier is the direct image of the usual $t\,e^{-t}$ resonance fix.

Common pitfall. Trying the ordinary particular solution $Ax^{-1}$ first and getting the contradiction $0=4x^{-1}$ without recognizing why — always check the forcing exponent against the indicial roots before choosing a trial.

Source. Kreyszig, Advanced Engineering Mathematics, Ch. 2 (§2.5 Euler–Cauchy Equations).

Question 2 of 8: Cauchy–Euler Equation with a Resonant Power-Law Forcing

Question 2: Cauchy–Euler Equation with a Resonant Power-Law Forcing (20 marks)