Euler–Cauchy Equation with Exponential–Polynomial Forcing: 04-BS-1, December 2016, Q1

Question 1: Euler–Cauchy Equation with Exponential–Polynomial 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 homogeneous (equidimensional) Euler–Cauchy operator on the left, forced by a polynomial-times-exponential right-hand side.

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

Approach. Solve the Euler–Cauchy homogeneous equation by trying $y=x^m$, then use variation of parameters (not undetermined coefficients, since the forcing is not compatible with the Euler operator's own eigenfunctions) to build a particular solution.

Homogeneous solution. Trying $y=x^m$ in $x^2y''-2xy'+2y=0$: $m(m-1)-2m+2=0\Rightarrow m^2-3m+2=0\Rightarrow(m-1)(m-2)=0\Rightarrow m=1,2$. So $$y_h=C_1x+C_2x^2,\qquad y_1=x,\ y_2=x^2.$$
Standard form and Wronskian. Dividing by $x^2$: $y''-\tfrac2xy'+\tfrac2{x^2}y=(1-2x)xe^{-2x}=g(x)$. The Wronskian is $$W=y_1y_2'-y_1'y_2=x(2x)-1\cdot x^2=x^2.$$
Variation-of-parameters integrands. $$u_1'=-\frac{y_2\,g}{W}=-\frac{x^2(1-2x)xe^{-2x}}{x^2}=(2x^2-x)e^{-2x},\qquad u_2'=\frac{y_1\,g}{W}=\frac{x(1-2x)xe^{-2x}}{x^2}=(1-2x)e^{-2x}.$$
Integrate. Using $\int x^ne^{-2x}dx$ by parts repeatedly, $$u_2=\int(1-2x)e^{-2x}dx=xe^{-2x},\qquad u_1=\int(2x^2-x)e^{-2x}dx=-e^{-2x}\!\left(x^2+\tfrac{x}{2}+\tfrac14\right)$$ (constants of integration dropped — any particular antiderivative suffices for a particular solution).
Assemble $y_p=u_1y_1+u_2y_2$. $$y_p=-e^{-2x}\!\left(x^2+\tfrac{x}{2}+\tfrac14\right)x+xe^{-2x}\cdot x^2=e^{-2x}\!\left[-x^3-\tfrac{x^2}{2}-\tfrac{x}{4}+x^3\right]=-\left(\tfrac{x^2}{2}+\tfrac{x}{4}\right)e^{-2x}.$$ General solution: $$y(x)=\boxed{C_1x+C_2x^2-\left(\dfrac{x^2}{2}+\dfrac{x}{4}\right)e^{-2x}}$$ (Check: substituting $y_p$ back reproduces $x^3(1-2x)e^{-2x}$ exactly.)

Quantity	Result
Homogeneous roots	$m=1,2$
Particular solution $y_p$	$-\left(\tfrac{x^2}{2}+\tfrac{x}{4}\right)e^{-2x}$
General solution $y(x)$	$C_1x+C_2x^2-\left(\tfrac{x^2}{2}+\tfrac{x}{4}\right)e^{-2x}$

Concept & Theory

Topic. Euler–Cauchy (equidimensional) second-order ODEs solved by variation of parameters.

Key relations. $ax^2y''+bxy'+cy=0$ has solutions $y=x^m$ with $m$ satisfying $am(m-1)+bm+c=0$. For non-Euler-compatible forcing, variation of parameters gives $y_p=u_1y_1+u_2y_2$ with $u_1'=-y_2g/W$, $u_2'=y_1g/W$ (standard form $g(x)$, i.e. the RHS after dividing by the leading coefficient $x^2$).

Why this works. The forcing $(1-2x)x^3e^{-2x}$ is not itself annihilated by any constant-coefficient differential operator compatible with the $x^m$ basis, so undetermined coefficients has no clean trial form here; but dividing by the Wronskian $W=x^2$ exactly cancels the $x^2$ already in the standard-form RHS, leaving a pure polynomial-times-exponential integrand for both $u_1',u_2'$ — always elementary to integrate by parts.

Common pitfall. Forgetting to convert to standard form (dividing by $x^2$) before reading off $g(x)$ for the variation-of-parameters formula; sign errors in $u_1'=-y_2g/W$ (the minus sign is easy to drop); attempting undetermined coefficients directly against an Euler-Cauchy operator, which has no constant-coefficient trial-function theory.

Source. Kreyszig, Advanced Engineering Mathematics, Ch. 2 (§2.5 Euler–Cauchy Equations, §2.10 Variation of Parameters).

Question 1 of 8: Euler–Cauchy Equation with Exponential–Polynomial Forcing

Question 1: Euler–Cauchy Equation with Exponential–Polynomial Forcing 20 marks