General Solutions of Three First–/Second–Order ODEs: 04-BS-1, Undated paper, Q1

Question 1: General Solutions of Three First–/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 independent first-order-reducible ODEs in $x$: (a) a constant-coefficient second-order equation missing $y$, forced by $\sin2x$; (b) a separable (Bernoulli, $n=2$) equation; (c) a first-order linear equation.

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

Approach. (a) Solve the homogeneous equation via its characteristic roots, then use undetermined coefficients for the sinusoidal forcing. (b) Separate variables directly. (c) Divide through to standard linear form and apply an integrating factor.

(a) Homogeneous solution. The characteristic equation $r^2+2r=0$ factors as $r(r+2)=0$, giving $r=0,-2$: $$y_h=C_1+C_2e^{-2x}.$$
(a) Particular solution. Try $y_p=A\cos2x+B\sin2x$. Then $y_p''+2y_p'=(-4A+4B)\cos2x+(-4A-4B)\sin2x$. Matching to $\sin2x$: $-4A+4B=0$ and $-4A-4B=1$, so $A=B=-\tfrac18$: $$y_p=-\tfrac18\cos2x-\tfrac18\sin2x.$$ $$y(x)=\boxed{C_1+C_2e^{-2x}-\tfrac18\cos2x-\tfrac18\sin2x}$$
(b) Separate variables. $y'=-xy^2\Rightarrow\dfrac{dy}{y^2}=-x\,dx\Rightarrow-\dfrac1y=-\dfrac{x^2}{2}+C_0$: $$y(x)=\boxed{\dfrac{2}{x^2+C}}\quad(\text{plus the singular solution }y\equiv0).$$
(c) Standard linear form and integrating factor. Divide by $2x$: $y'+\dfrac{1}{2x}y=x^{1/2}$. The integrating factor is $\mu=e^{\int\frac{1}{2x}dx}=x^{1/2}$, and $\big(x^{1/2}y\big)'=x^{1/2}\cdot x^{1/2}=x$.
(c) Integrate. $x^{1/2}y=\dfrac{x^2}{2}+C\ \Rightarrow$ $$y(x)=\boxed{\tfrac12x^{3/2}+Cx^{-1/2}}$$

Part	General solution
(a)	$y=C_1+C_2e^{-2x}-\tfrac18(\cos2x+\sin2x)$
(b)	$y=2/(x^2+C)$ (plus $y\equiv0$)
(c)	$y=\tfrac12x^{3/2}+Cx^{-1/2}$

Concept & Theory

Topic. Reducible/separable/linear first-order techniques and undetermined coefficients for a $y$-free second-order ODE.

Key relations. A second-order ODE with no $y$ term reduces to first order in $u=y'$; for $y'+P(x)y=Q(x)$, the integrating factor $\mu=e^{\int P\,dx}$ makes $(\mu y)'=\mu Q$; separable equations integrate directly after dividing by the $y$-dependent factor.

Why this works. In (a), because $r=0$ is a root, the forcing $\sin2x$ (frequency $2$, not matching the imaginary part of any root) needs only the ordinary trial $A\cos2x+B\sin2x$, no resonance correction. In (c), $1/(2x)$ integrates to $\tfrac12\ln x$, giving the clean power-law integrating factor $x^{1/2}$.

Common pitfall. In (b), dividing by $y^2$ silently discards the constant solution $y\equiv0$ — it must be reported alongside the family. In (a), forgetting that $r=0$ means $C_1$ (not $C_1e^{0\cdot x}$) is already the full constant mode.

Source. Kreyszig, Advanced Engineering Mathematics, Ch. 1 (§1.3 Separable ODEs, §1.5 Linear ODEs) and Ch. 2 (§2.7–2.8 Undetermined Coefficients).

Question 1 of 7: General Solutions of Three First–/Second–Order ODEs

Question 1: General Solutions of Three First–/Second–Order ODEs (a) 7, (b) 7, (c) 6 marks