Two Initial Value Problems — Separable and Constant-Coefficient: 04-BS-1, May 2018, Q2

Question 2: Two Initial Value Problems — Separable and Constant-Coefficient (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) The separable first-order IVP $y'+2ty^2=0,\ y(1)=2$. (b) The linear constant-coefficient second-order IVP $y''-y'-2y=3t^2,\ y(0)=0,\ y'(0)=0$.

Find. The unique solution $y(t)$ in each case.

Approach. (a) Separate variables and integrate, then fix the constant from $y(1)=2$. (b) Solve the homogeneous equation (real distinct roots), find a quadratic particular solution by undetermined coefficients, then fix both constants from the zero initial conditions.

(a) Separate and integrate. $\dfrac{dy}{y^2}=-2t\,dt\ \Rightarrow\ -\dfrac1y=-t^2+C\ \Rightarrow\ \dfrac1y=t^2-C$. Writing $\dfrac1y=t^2+K$ and applying $y(1)=2$: $\tfrac12=1+K\Rightarrow K=-\tfrac12$, so $$y(t)=\frac1{t^2-\tfrac12}=\frac2{2t^2-1}.$$
(b) Homogeneous solution. Characteristic equation $r^2-r-2=0=(r-2)(r+1)\Rightarrow r=2,-1$, so $$y_h(t)=C_1e^{2t}+C_2e^{-t}.$$
(b) Particular solution. Try $y_p=At^2+Bt+C$: $y_p'=2At+B,\ y_p''=2A$. Substituting into $y''-y'-2y=3t^2$: $$2A-(2At+B)-2(At^2+Bt+C)=-2At^2+(-2A-2B)t+(2A-B-2C)=3t^2.$$ Matching coefficients: $-2A=3\Rightarrow A=-\tfrac32$; $-2A-2B=0\Rightarrow B=\tfrac32$; $2A-B-2C=0\Rightarrow C=-\tfrac94$. So $$y_p(t)=-\tfrac32t^2+\tfrac32t-\tfrac94.$$
(b) Apply the initial conditions. $y(t)=C_1e^{2t}+C_2e^{-t}-\tfrac32t^2+\tfrac32t-\tfrac94$. $y(0)=C_1+C_2-\tfrac94=0\Rightarrow C_1+C_2=\tfrac94$. $y'(t)=2C_1e^{2t}-C_2e^{-t}-3t+\tfrac32$, so $y'(0)=2C_1-C_2+\tfrac32=0\Rightarrow2C_1-C_2=-\tfrac32$. Adding: $3C_1=\tfrac34\Rightarrow C_1=\tfrac14$, then $C_2=2$.

$$\text{(a)}\quad y(t)=\boxed{\frac2{2t^2-1}}\qquad\qquad\text{(b)}\quad y(t)=\boxed{\tfrac14e^{2t}+2e^{-t}-\tfrac32t^2+\tfrac32t-\tfrac94}$$

Quantity	Result
(a) $y(t)$	$2/(2t^2-1)$
(b) Homogeneous roots	$r=2,-1$
(b) Particular solution	$-\tfrac32t^2+\tfrac32t-\tfrac94$
(b) $C_1,C_2$	$\tfrac14,\ 2$

Concept & Theory

Topic. Separable first-order ODEs and undetermined coefficients for constant-coefficient linear ODEs.

Key relations. Separable form $\dfrac{dy}{g(y)}=f(t)\,dt$ integrates directly. For $ay''+by'+cy=$ polynomial of degree $n$ (with $0$ not a root of the characteristic equation), try a matching-degree polynomial $y_p$.

Why this works. In (a), dividing by $y^2$ separates the variables cleanly since the equation has no other $y$-dependence. In (b), because neither characteristic root is $0$, a plain degree-2 polynomial trial (no resonance factor of $t$) is guaranteed to match a degree-2 forcing term.

Common pitfall. In (a), losing the singular solution $y\equiv0$ when dividing by $y^2$ (irrelevant here since $y(1)=2\ne0$, but worth noting). In (b), dropping the constant term $C$ in the polynomial trial, which is needed even though the forcing $3t^2$ has no constant term, because the first-derivative coupling $-y'$ generates one.

Source. Kreyszig, Advanced Engineering Mathematics, Ch. 1 (§1.3 Separable ODEs) and Ch. 2 (§2.7 Nonhomogeneous ODEs: Undetermined Coefficients).

Question 2 of 8: Two Initial Value Problems — Separable and Constant-Coefficient

Question 2: Two Initial Value Problems — Separable and Constant-Coefficient (a) 10, (b) 10 marks