Three First- and Second-Order ODEs — Integrating Factor, Separable/Bernoulli, and Complex Characteristic Roots: 04-BS-1, May 2015, Q1

Question 1: Three First- and Second-Order ODEs — Integrating Factor, Separable/Bernoulli, and Complex Characteristic Roots (a) 7, (b) 7, (c) 6 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. Three unrelated ODEs: (a) a linear first-order equation with an exponential forcing term shaped to match the integrating factor; (b) a separable (Bernoulli, $n=2$) first-order equation; (c) a constant-coefficient linear second-order homogeneous equation.

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

Approach. (a) Multiply by the integrating factor $\mu=e^{\int 2x\,dx}=e^{x^2}$ so the left side collapses to $(\mu y)'$. (b) Separate variables directly (equivalently, this is Bernoulli with $n=2$: dividing by $y^2$ linearizes it in $1/y$). (c) Solve the characteristic equation; complex roots give an oscillatory-times-exponential solution.

(a) Integrating factor. $\mu=e^{\int 2x\,dx}=e^{x^2}$. Multiplying through, $$\big(e^{x^2}y\big)'=e^{x^2}\big(y'+2xy\big)=e^{x^2}\cdot2xe^{-x^2}=2x.$$
(a) Integrate and solve for $y$. $e^{x^2}y=\displaystyle\int 2x\,dx=x^2+C_1$, so $$y(x)=\boxed{(x^2+C_1)e^{-x^2}}.$$
(b) Separate variables. For $y\ne0$, $\dfrac{dy}{y^2}=-2x\,dx$. Integrating both sides, $$-\frac1y=-x^2+C\ \Rightarrow\ \frac1y=x^2-C\ \Rightarrow\ y(x)=\boxed{\dfrac{1}{x^2+C_2}}$$ (renaming $-C\to C_2$); the division by $y^2$ also loses the singular solution $y\equiv0$, which independently satisfies the ODE.
(c) Characteristic equation. $r^2-2r+3=0\ \Rightarrow\ r=\dfrac{2\pm\sqrt{4-12}}{2}=1\pm i\sqrt2$. Complex roots $p\pm qi$ with $p=1,\ q=\sqrt2$ give $$y(x)=\boxed{e^{x}\big(C_3\cos(\sqrt2\,x)+C_4\sin(\sqrt2\,x)\big)}.$$

Part	Result
(a) $y(x)$	$(x^2+C_1)e^{-x^2}$
(b) $y(x)$	$\dfrac{1}{x^2+C_2}$ (plus the singular solution $y\equiv0$)
(c) $y(x)$	$e^{x}\big(C_3\cos(\sqrt2\,x)+C_4\sin(\sqrt2\,x)\big)$

Concept & Theory

Topic. First-order linear ODEs via integrating factors, separable/Bernoulli equations, and constant-coefficient second-order ODEs with complex roots.

Key relations. For $y'+P(x)y=Q(x)$, the integrating factor $\mu=e^{\int P\,dx}$ makes $(\mu y)'=\mu Q$. A separable equation $dy/f(y)=g(x)dx$ integrates directly. For $ay''+by'+cy=0$ with complex roots $r=p\pm qi$, $y=e^{px}(C_1\cos qx+C_2\sin qx)$.

Why this works. In (a), the forcing $2xe^{-x^2}$ was engineered so that multiplying by $\mu=e^{x^2}$ exactly cancels the exponential, leaving a trivial $\int2x\,dx$. In (c), a negative discriminant means the two exponential solutions $e^{r_1x},e^{r_2x}$ are complex conjugates; Euler's formula recombines them into real oscillatory form.

Common pitfall. In (b), dividing by $y^2$ silently discards the constant solution $y\equiv0$ — it must be reported alongside the family $1/(x^2+C_2)$. In (c), forgetting the exponential envelope $e^{x}$ (writing only the trig part) misses that the solution is not periodic.

Source. Kreyszig, Advanced Engineering Mathematics, Ch. 1 (§1.5 Linear ODEs, Integrating Factor; §1.3 Separable ODEs) and Ch. 2 (§2.2 Homogeneous Linear ODEs of Second Order).

Question 1 of 8: Three First- and Second-Order ODEs — Integrating Factor, Separable/Bernoulli, and Complex Characteristic Roots

Question 1: Three First- and Second-Order ODEs — Integrating Factor, Separable/Bernoulli, and Complex Characteristic Roots (a) 7, (b) 7, (c) 6 marks