Euler–Cauchy Equation with Non-Resonant Exponential–Polynomial Forcing: 04-BS-1, December 2017, Q2

Question 2: Euler–Cauchy Equation with Non-Resonant 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. An Euler–Cauchy (equidimensional) ODE with forcing $(1-2x)x^3e^{-2x}$, which is not itself an Euler-compatible power of $x$ (it carries an $e^{-2x}$ factor), so undetermined coefficients does not apply directly.

Find. The general solution $y(x)$ (two arbitrary constants).

Approach. Solve the homogeneous Euler–Cauchy equation with the power trial $y=x^m$, then build a particular solution by variation of parameters using the two homogeneous solutions as a basis. Dividing the forcing by the Wronskian is engineered by the source paper to cancel the denominator exactly, leaving an elementary polynomial×exponential integral.

Homogeneous solution. Try $y=x^m$: $x^2\cdot m(m-1)x^{m-2}-2x\cdot mx^{m-1}+2x^m=0\Rightarrow m^2-3m+2=0=(m-1)(m-2)$, so $m=1,2$ and $$y_h(x)=C_1x+C_2x^2.$$
Standard form and Wronskian. Dividing the ODE by $x^2$ puts it in standard form $y''-\tfrac2xy'+\tfrac2{x^2}y=g(x)$ with $g(x)=\dfrac{(1-2x)x^3e^{-2x}}{x^2}=(1-2x)xe^{-2x}$. With $y_1=x,\ y_2=x^2$, the Wronskian is $$W=y_1y_2'-y_2y_1'=x(2x)-x^2(1)=x^2.$$
Variation-of-parameters integrals. The formula $y_p=-y_1\displaystyle\int\frac{y_2g}{W}dx+y_2\int\frac{y_1g}{W}dx$ needs $$\frac{y_2g}{W}=\frac{x^2(1-2x)xe^{-2x}}{x^2}=(x-2x^2)e^{-2x},\qquad \frac{y_1g}{W}=\frac{x(1-2x)xe^{-2x}}{x^2}=(1-2x)e^{-2x}$$ — the $x^2$ in $W$ exactly cancels the $x^2$ in the numerators, leaving elementary polynomial×exponential integrands, both done by parts: $$\int(1-2x)e^{-2x}dx=xe^{-2x},\qquad \int(x-2x^2)e^{-2x}dx=\left(x^2+\tfrac x2+\tfrac14\right)e^{-2x}.$$
Assemble the particular solution. $$y_p=-x\left(x^2+\tfrac x2+\tfrac14\right)e^{-2x}+x^2\left(xe^{-2x}\right)=\left(-x^3-\tfrac{x^2}2-\tfrac x4+x^3\right)e^{-2x}=-\left(\frac{x^2}2+\frac x4\right)e^{-2x}.$$

$$y(x)=\boxed{C_1x+C_2x^2-\left(\frac{x^2}2+\frac x4\right)e^{-2x}}$$

Quantity	Result
Homogeneous solution	$C_1x+C_2x^2$
Wronskian	$W=x^2$
Particular solution	$-\left(\tfrac{x^2}2+\tfrac x4\right)e^{-2x}$

Concept & Theory

Topic. Variation of parameters for a nonhomogeneous Euler–Cauchy equation.

Key relations. For $y''+p(x)y'+q(x)y=g(x)$ with homogeneous basis $y_1,y_2$ and Wronskian $W=y_1y_2'-y_2y_1'$, the particular solution is $y_p=-y_1\int\frac{y_2g}{W}dx+y_2\int\frac{y_1g}{W}dx$. This applies whenever the forcing is not a form undetermined coefficients can match (here, an exponential times a non-Euler power of $x$).

Why this works. The equation must first be divided by the leading coefficient $x^2$ to reach standard form before $W$ and $g$ are defined; for a 2nd-order Euler–Cauchy equation with roots $m_1,m_2$, $W$ is always a power of $x$ (here $x^{m_1+m_2-1}=x^2$), and here it cancels the $x^2$ already present in the forcing numerator, leaving a purely elementary integrand that a naive undetermined-coefficients guess (needing an Euler-compatible power) could never reach.

Common pitfall. Forgetting to divide the original ODE by $x^2$ before extracting $g(x)$ — using the un-normalized RHS $(1-2x)x^3e^{-2x}$ directly in the variation-of-parameters formula gives a result off by a factor of $x^2$. Also sign errors in the by-parts reduction of $\int x^2e^{-2x}dx$.

Source. Kreyszig, Advanced Engineering Mathematics, Ch. 1 (§1.6 Euler–Cauchy Equations) and Ch. 2 (§2.10 Variation of Parameters).

Question 2 of 8: Euler–Cauchy Equation with Non-Resonant Exponential–Polynomial Forcing

Question 2: Euler–Cauchy Equation with Non-Resonant Exponential–Polynomial Forcing (20 marks)