Three First- and Second-Order ODEs: 04-BS-1, May 2016, 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. (a) A first-order linear ODE whose left side is already a disguised exact derivative. (b) A separable, nonlinear first-order ODE. (c) A homogeneous Euler–Cauchy (equidimensional) second-order ODE.

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

Approach. (a) Recognize $xy'+y=(xy)'$ and integrate directly. (b) Separate variables. (c) Try $y=x^m$ to reduce the Euler–Cauchy equation to an algebraic indicial equation.

(a) Collapse the left side to an exact derivative. By the product rule, $\dfrac{d}{dx}(xy)=xy'+y$, so the ODE is exactly $$(xy)'=2\cos(3x).$$
(a) Integrate and solve for $y$. Integrating both sides, $$xy=\int 2\cos(3x)\,dx=\tfrac23\sin(3x)+C.$$ Dividing by $x$, $$y(x)=\boxed{\dfrac{2\sin(3x)}{3x}+\dfrac{C}{x}}$$ (Check: $xy=\tfrac23\sin3x+C\Rightarrow(xy)'=2\cos3x$, and $(xy)'=xy'+y$ by the product rule ✓.)
(b) Separate variables. $y'+2xy^2=0\Rightarrow \dfrac{dy}{y^2}=-2x\,dx$ (excluding the trivial solution $y\equiv0$, which also satisfies the ODE). Integrating, $$-\frac1y=-x^2+K \;\Rightarrow\; \frac1y=x^2-K.$$
(b) Solve for $y$. Relabelling the constant ($C=-K$), $$y(x)=\boxed{\dfrac{1}{x^2+C}}$$ (Check: $y'=-\dfrac{2x}{(x^2+C)^2}$, and $2xy^2=\dfrac{2x}{(x^2+C)^2}$, so $y'+2xy^2=0$ ✓.)
(c) Euler–Cauchy trial solution. For $2x^2y''+5xy'-2y=0$, try $y=x^m$: then $y'=mx^{m-1}$, $y''=m(m-1)x^{m-2}$, and substituting gives $$2m(m-1)x^m+5mx^m-2x^m=0 \;\Rightarrow\; 2m^2+3m-2=0.$$
(c) Solve the indicial equation. By the quadratic formula, $$m=\frac{-3\pm\sqrt{9+16}}{4}=\frac{-3\pm5}{4}\;\Rightarrow\; m=\tfrac12,\ -2.$$ Two distinct real roots give two independent power-law modes: $$y(x)=\boxed{C_1\sqrt{x}+\dfrac{C_2}{x^2}}$$ (Check at $m=\tfrac12$: $2x^2\!\left(-\tfrac14x^{-3/2}\right)+5x\!\left(\tfrac12x^{-1/2}\right)-2x^{1/2}=-\tfrac12x^{1/2}+\tfrac52x^{1/2}-2x^{1/2}=0$ ✓.)

Part	Result
(a) $y(x)$	$\dfrac{2\sin(3x)}{3x}+\dfrac{C}{x}$
(b) $y(x)$	$\dfrac{1}{x^2+C}$
(c) $y(x)$	$C_1\sqrt{x}+\dfrac{C_2}{x^2}$

Concept & Theory

Topic. Exact/linear first-order ODEs, separable ODEs, and the Euler–Cauchy (equidimensional) second-order ODE.

Key relations. An ODE of the form $xy'+y=f(x)$ is always exactly $(xy)'=f(x)$ by the product rule — no integrating factor is needed. A separable ODE $g(y)y'=f(x)$ integrates as $\int g(y)\,dy=\int f(x)\,dx$. The Euler–Cauchy equation $ax^2y''+bxy'+cy=0$ admits power solutions $y=x^m$, reducing to the indicial equation $am(m-1)+bm+c=0$.

Why this works. In (a), the specific combination $xy'+y$ is recognizable as $d(xy)/dx$, so the equation is already in exact form. In (c), each term $x^ky^{(k)}$ produces exactly one power of $x^m$ under $y=x^m$ (the $x$-powers and derivative-order reductions cancel), which is why Euler–Cauchy equations always collapse to a polynomial in $m$.

Common pitfall. In (a), missing that $xy'+y$ is already an exact derivative and instead applying a needless integrating factor $e^{\int dx/x}=x$ (which happens to also work, but is redundant). In (b), forgetting the sign flip when inverting $-1/y=-x^2+K$. In (c), sign errors in the indicial equation $2m^2+3m-2=0$, e.g. mixing up $+3m$ and $-3m$.

Source. Kreyszig, Advanced Engineering Mathematics, Ch. 1 (§1.3 Separable ODEs, §1.5 Linear ODEs) and Ch. 2 (§2.5 Euler–Cauchy Equations).

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