Cauchy–Euler Equation with a Resonant Power-Law Forcing: 04-BS-1, May 2015, 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 $2x^2y''-5xy'-4y=3x^4$, $x>0$.

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

Approach. Try $y=x^m$ for the homogeneous equation to get the indicial polynomial. Check the forcing exponent against the indicial roots — if it matches one, the ordinary trial $Ax^4$ fails and must be replaced by the resonant Cauchy–Euler trial $Ax^4\ln x$.

Indicial equation. Substituting $y=x^m$ into $2x^2y''-5xy'-4y=0$: $2m(m-1)-5m-4=0\Rightarrow2m^2-7m-4=0=(2m+1)(m-4)$, so $m=4,\ -\tfrac12$. $$y_h=C_1x^4+C_2x^{-1/2}.$$
Resonance check. The forcing $3x^4$ matches the homogeneous mode $x^4$ ($m=4$), so $y_p=Ax^4$ would solve to $0=3x^4$ — instead use $y_p=Ax^4\ln x$.
Differentiate the trial. $$y_p'=Ax^3(4\ln x+1),\qquad y_p''=Ax^2\big(12\ln x+7\big).$$
Substitute and solve for $A$. $$2x^2y_p''-5xy_p'-4y_p=Ax^4\big[2(12\ln x+7)-5(4\ln x+1)-4\ln x\big]=Ax^4\big[(24-20-4)\ln x+(14-5)\big]=9Ax^4.$$ Setting $9A=3$ gives $A=\tfrac13$. $$y_p=\tfrac13x^4\ln x.$$

$$y(x)=\boxed{C_1x^4+C_2x^{-1/2}+\tfrac13x^4\ln x}$$

Quantity	Result
Indicial roots	$m=4,\ -1/2$
Particular solution	$\tfrac13x^4\ln x$
General solution	$y(x)=C_1x^4+C_2x^{-1/2}+\tfrac13x^4\ln x$

Concept & Theory

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

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

Why this works. The substitution $x=e^t$ turns a Cauchy–Euler equation into a constant-coefficient ODE in $t$; the power-law forcing $x^4$ becomes $e^{4t}$ there, exactly a resonant exponential forcing whose fix is the familiar $t\,e^{4t}$ term — which is $\ln x\cdot x^4$ back in the original variable.

Common pitfall. Trying the ordinary particular solution $Ax^4$ first and reaching the contradiction $0=3x^4$ without recognizing why — always compare 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)