Two First- and Second-Order ODEs: 04-BS-1, December 2015, Q1

Question 1: Two First- and Second-Order ODEs (a) 10, (b) 10 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) A first-order linear ODE with variable coefficient $x$ and forcing $2xe^{-x^2}$. (b) A homogeneous, constant-coefficient second-order ODE.

Find. The general solution $y(x)$ in each case.

Approach. (a) Solve by an integrating factor $\mu(x)=e^{\int x\,dx}$. (b) Solve the characteristic (auxiliary) equation and combine the two real exponential modes.

(a) Integrating factor. The equation $y'+xy=2xe^{-x^2}$ has integrating factor $$\mu(x)=e^{\int x\,dx}=e^{x^2/2}.$$
(a) Multiply through and integrate. Multiplying both sides by $\mu$, $$\left(e^{x^2/2}y\right)'=2xe^{-x^2}\cdot e^{x^2/2}=2xe^{-x^2/2}.$$ The right side integrates directly, since $\dfrac{d}{dx}e^{-x^2/2}=-xe^{-x^2/2}$: $$\int 2xe^{-x^2/2}\,dx=-2e^{-x^2/2}+C.$$ So $e^{x^2/2}y=-2e^{-x^2/2}+C$.
(a) Solve for $y$. Dividing by $e^{x^2/2}$, $$y(x)=\boxed{Ce^{-x^2/2}-2e^{-x^2}}$$ (Check: $y'=-Cxe^{-x^2/2}+4xe^{-x^2}$, and $y'+xy=-Cxe^{-x^2/2}+4xe^{-x^2}+Cxe^{-x^2/2}-2xe^{-x^2}=2xe^{-x^2}$ ✓.)
(b) Characteristic equation. For $y''+y'-6y=0$, try $y=e^{rx}$: $$r^2+r-6=0=(r+3)(r-2)=0 \;\Rightarrow\; r=-3,\ 2.$$
(b) General solution. Two distinct real roots give two independent exponential modes: $$y(x)=\boxed{C_1e^{-3x}+C_2e^{2x}}$$

Part	Result
(a) $y(x)$	$Ce^{-x^2/2}-2e^{-x^2}$
(b) $y(x)$	$C_1e^{-3x}+C_2e^{2x}$

Concept & Theory

Topic. Integrating factors for first-order linear ODEs; characteristic equations for constant-coefficient homogeneous ODEs.

Key relations. For $y'+p(x)y=q(x)$, the integrating factor $\mu=e^{\int p\,dx}$ makes $(\mu y)'=\mu q$, which integrates directly. For $ay''+by'+cy=0$, substituting $y=e^{rx}$ gives the characteristic equation $ar^2+br+c=0$; distinct real roots $r_1,r_2$ give $y=C_1e^{r_1x}+C_2e^{r_2x}$.

Why this works. In (a), $\mu=e^{x^2/2}$ is chosen precisely so that $\mu'=x\mu$, making the left side collapse to the exact derivative $(\mu y)'$ by the product rule. In (b), $e^{rx}$ is an eigenfunction of the differentiation operator, so the ODE becomes purely algebraic in $r$.

Common pitfall. In (a), forgetting the factor of $\tfrac12$ inside the exponent of $\mu$ (integrating factor is $e^{x^2/2}$, not $e^{x^2}$), which then requires recognizing $2xe^{-x^2/2}$ (not $2xe^{-x^2}$) as the right side after multiplying through. In (b), sign errors factoring $r^2+r-6$.

Source. Kreyszig, Advanced Engineering Mathematics, Ch. 1 (§1.5 Linear ODEs, Integrating Factor) and Ch. 2 (§2.2 Homogeneous Linear ODEs of Second Order).

Question 1 of 8: Two First- and Second-Order ODEs

Question 1: Two First- and Second-Order ODEs (a) 10, (b) 10 marks