Two Initial Value Problems: 04-BS-1, December 2015, Q2

Question 2: Two Initial Value Problems (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 separable, nonlinear first-order ODE with $y(1)=2$. (b) A constant-coefficient linear ODE with complex characteristic roots, forced at $\cos(3t)$, and zero initial conditions.

Find. $y(t)$ satisfying each IVP.

Approach. (a) Separate variables and integrate directly. (b) Solve the homogeneous equation (complex roots), find a particular solution by undetermined coefficients, then fix the two constants from the ICs.

(a) Separate and integrate. $\dfrac{dy}{y^2}=-2t\,dt \;\Rightarrow\; -\dfrac1y=-t^2+K \;\Rightarrow\; \dfrac1y=t^2+C.$
(a) Apply the initial condition. $y(1)=2\Rightarrow\dfrac12=1+C\Rightarrow C=-\dfrac12$. So $$y(t)=\frac{1}{t^2-\tfrac12}=\boxed{\dfrac{2}{2t^2-1}}$$ (Check: $y(1)=2/(2-1)=2$ ✓; $y'=-8t/(2t^2-1)^2$ and $2ty^2=8t/(2t^2-1)^2$, so $y'+2ty^2=0$ ✓.)
(b) Homogeneous solution. $r^2-12r+45=0\Rightarrow r=\dfrac{12\pm\sqrt{144-180}}{2}=6\pm3i$, so $$y_h(t)=e^{6t}(C_1\cos3t+C_2\sin3t).$$
(b) Particular solution. Try $y_p=A\cos3t+B\sin3t$ (no resonance, since the homogeneous modes carry the $e^{6t}$ envelope). Substituting and matching coefficients of $\cos3t$ and $\sin3t$ against $18\cos3t+0\sin3t$: $$36A-36B=18,\qquad 36A+36B=0.$$ The second equation gives $B=-A$; substituting into the first, $72A=18\Rightarrow A=\tfrac14,\ B=-\tfrac14$. So $y_p=\tfrac14\cos3t-\tfrac14\sin3t$.
(b) Apply the initial conditions. $y(t)=e^{6t}(C_1\cos3t+C_2\sin3t)+\tfrac14\cos3t-\tfrac14\sin3t$. $y(0)=C_1+\tfrac14=0\Rightarrow C_1=-\tfrac14$. Differentiating, $y'(t)=e^{6t}\big[(6C_1+3C_2)\cos3t+(6C_2-3C_1)\sin3t\big]-\tfrac34\sin3t-\tfrac34\cos3t$, so $$y'(0)=6C_1+3C_2-\tfrac34=0 \;\Rightarrow\; 3C_2=\tfrac34-6\left(-\tfrac14\right)=\tfrac34+\tfrac32=\tfrac94 \;\Rightarrow\; C_2=\tfrac34.$$

$$y(t)=\boxed{e^{6t}\left(-\tfrac14\cos3t+\tfrac34\sin3t\right)+\tfrac14\cos3t-\tfrac14\sin3t}$$

Part	Result
(a) $y(t)$	$\dfrac{2}{2t^2-1}$
(b) $y(t)$	$e^{6t}\left(-\tfrac14\cos3t+\tfrac34\sin3t\right)+\tfrac14\cos3t-\tfrac14\sin3t$

Concept & Theory

Topic. Separable nonlinear ODEs; underdamped-form forced linear ODEs with complex characteristic roots.

Key relations. A separable ODE $dy/dt=f(t)g(y)$ integrates as $\int dy/g(y)=\int f(t)\,dt$. For $ay''+by'+cy=F_0\cos\omega t$ with complex roots $\alpha\pm i\beta$ (and $\beta\ne\omega$, no resonance), the steady-state particular solution has the same frequency: $y_p=A\cos\omega t+B\sin\omega t$, found by direct substitution.

Why this works. In (a), $y^{-2}$ has a clean antiderivative ($-1/y$), so separating variables converts the ODE to an algebraic relation between $t$ and $y$. In (b), $\alpha=6\ne0$ means $y_h$ grows/decays exponentially rather than resonating with the bounded $\cos3t$ forcing, so the un-multiplied trial $A\cos3t+B\sin3t$ is linearly independent of $y_h$ and works directly.

Common pitfall. In (a), losing the constant of integration or mishandling the sign when inverting $1/y=t^2+C$. In (b), arithmetic slips solving the $2\times2$ system for $A,B$, or dropping the product-rule cross term $6C_1+3C_2$ when differentiating $e^{6t}(C_1\cos3t+C_2\sin3t)$.

Source. Kreyszig, Advanced Engineering Mathematics, Ch. 1 (§1.3 Separable ODEs) and Ch. 2 (§2.8 Modeling: Forced Oscillations, Resonance).

Question 2 of 8: Two Initial Value Problems

Question 2: Two Initial Value Problems (a) 10, (b) 10 marks