Euler–Cauchy Equation with a Non-Resonant Power Forcing: 04-BS-1, May 2017, Q2

Question 2: Euler–Cauchy Equation with a Non-Resonant Power 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. An Euler–Cauchy (equidimensional) equation forced by a pure power $4x^{-2}$.

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

Approach. Solve the homogeneous Euler–Cauchy equation by trying $y=x^m$, then check whether the forcing power $x^{-2}$ coincides with a homogeneous root (it does not), so an undetermined-coefficients power trial $y_p=Ax^{-2}$ applies directly.

Homogeneous solution. Trying $y=x^m$ in $2x^2y''+xy'-3y=0$: $2m(m-1)+m-3=0\Rightarrow2m^2-m-3=0\Rightarrow(2m-3)(m+1)=0\Rightarrow m=\tfrac32,\,-1$. So $$y_h=C_1x^{3/2}+C_2x^{-1}.$$
Check the forcing power against the roots. The forcing is $4x^{-2}$, i.e. power $-2$; since $-2\ne\tfrac32$ and $-2\ne-1$, the trial $y_p=Ax^{-2}$ is not a homogeneous solution and needs no resonance modification.
Substitute the trial. With $y_p=Ax^{-2}$: $y_p'=-2Ax^{-3}$, $y_p''=6Ax^{-4}$. $$2x^2(6Ax^{-4})+x(-2Ax^{-3})-3(Ax^{-2})=12Ax^{-2}-2Ax^{-2}-3Ax^{-2}=7Ax^{-2}.$$ Setting this equal to $4x^{-2}$ gives $A=\tfrac47$.
Assemble the general solution. $$\boxed{y(x)=C_1x^{3/2}+C_2x^{-1}+\dfrac47x^{-2}}$$ (Check: substituting $y_p=\tfrac47x^{-2}$ back reproduces $4x^{-2}$ exactly.)

Quantity	Result
Homogeneous roots	$m=\tfrac32,\,-1$
Particular solution $y_p$	$\tfrac47x^{-2}$
General solution $y(x)$	$C_1x^{3/2}+C_2x^{-1}+\tfrac47x^{-2}$

Concept & Theory

Topic. Euler–Cauchy (equidimensional) second-order ODEs with a power-law forcing, solved by undetermined coefficients.

Key relations. $ax^2y''+bxy'+cy=0$ has solutions $y=x^m$ with $m$ satisfying $am(m-1)+bm+c=0$. When the RHS is $Kx^k$ and $k$ is not a homogeneous root, the trial $y_p=Ax^k$ works directly, since every term of $2x^2y_p''+xy_p'-3y_p$ collapses to a multiple of $x^k$.

Why this works. Because $x^m$-type substitutions turn every term into a pure power of $x$, an Euler–Cauchy operator maps the trial $Ax^k$ onto exactly $[2k(k-1)+k-3]Ax^k$ — a single algebraic equation for $A$, with no differential terms of mismatched power to balance, provided $k$ isn't already a root (which would force a resonant $\ln x$ correction, as in the variation-of-parameters case seen in other 04-BS-1 papers).

Common pitfall. Not checking the forcing power against the homogeneous roots first (a resonant power needs an $x^k\ln x$ trial, not $Ax^k$); sign errors when collecting the coefficient of $Ax^{-2}$ across the three terms.

Source. Kreyszig, Advanced Engineering Mathematics, Ch. 2 (§2.5 Euler–Cauchy Equations, §2.7 Nonhomogeneous ODEs).

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

Question 2: Euler–Cauchy Equation with a Non-Resonant Power Forcing 20 marks