Three First- and Second-Order ODEs: 04-BS-1, May 2017, Q1

Question 1: Three First- and Second-Order ODEs (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 first- and second-order ODEs, each solvable by a different elementary technique.

Find. The general solution of each.

Approach. (a) Recognize the left side as an exact derivative $(x^2y)'$; (b) separate variables (a Bernoulli equation with $n=2$); (c) solve the constant-coefficient characteristic equation.

(a) Recognize the exact-derivative form. Since $\dfrac{d}{dx}(x^2y)=x^2y'+2xy$, the left side of $x^2y'+2xy=2\sin(3x)$ is already $\dfrac{d}{dx}(x^2y)$ — no integrating factor needs to be derived, it falls out directly from the product rule.
(a) Integrate both sides. $$x^2y=\int2\sin(3x)\,dx=-\tfrac23\cos(3x)+C\ \Rightarrow\ \boxed{y(x)=\dfrac{C-\tfrac23\cos(3x)}{x^2}}$$ (Check: substituting back reproduces $2\sin3x$ exactly.)
(b) Separate variables. $y'+2xy^2=0$ is Bernoulli with $n=2$; for $y\ne0$, divide by $y^2$: $\dfrac{dy}{y^2}=-2x\,dx$. Integrating, $-\dfrac1y=-x^2+C_1$, so $$\boxed{y(x)=\dfrac{1}{x^2+C}}$$ (with $C=-C_1$). The division by $y^2$ also discards the singular solution $y\equiv0$, which independently satisfies the equation.
(c) Characteristic equation. For $3y''+5y'-2y=0$, try $y=e^{rx}$: $3r^2+5r-2=0\Rightarrow r=\dfrac{-5\pm\sqrt{25+24}}{6}=\dfrac{-5\pm7}{6}$, giving $r=\tfrac13,\,-2$. So $$\boxed{y(x)=C_1e^{x/3}+C_2e^{-2x}}$$

Part	General solution
(a)	$y=\left(C-\tfrac23\cos3x\right)/x^2$
(b)	$y=1/(x^2+C)$ (plus singular $y\equiv0$)
(c)	$y=C_1e^{x/3}+C_2e^{-2x}$

Concept & Theory

Topic. Exact-derivative recognition, Bernoulli/separable equations, and constant-coefficient characteristic equations.

Key relations. An equation $P(x)y'+P'(x)y=g(x)$ is always $\big(P(x)y\big)'=g(x)$ by the product rule. A Bernoulli equation $y'+p(x)y=q(x)y^n$ reduces to separable form directly when $n=2$ and $p\equiv0$. A constant-coefficient linear ODE $ay''+by'+cy=0$ has solutions built from the roots of $ar^2+br+c=0$.

Why this works. Part (a) is deliberately disguised: recognizing $x^2y'+2xy$ as $(x^2y)'$ before reaching for the standard integrating-factor formula $\mu=e^{\int2/x\,dx}=x^2$ saves a step, though both routes agree. Part (b) has $n=2$, so $y^{-2}$ collapses the equation to a linear one in $u=1/y$ instantly.

Common pitfall. In (a), missing the shortcut and grinding through the full integrating-factor derivation (not wrong, just slower); in (b), dividing by $y^2$ without noting the lost solution $y\equiv0$; in (c), sign errors in the quadratic formula when the discriminant and leading coefficients are not both integers.

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

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

Question 1: Three First- and Second-Order ODEs (a) 7, (b) 7, (c) 6 marks