Two Linear ODEs — Resonant IVP and a Reducible Second-Order Equation: 04-BS-1, December 2013, Q1

Question 1: Two Linear ODEs — Resonant IVP and a Reducible Second-Order Equation (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 linear, constant-coefficient, nonhomogeneous ODE with forcing $2e^{-2t}$ and ICs $y(0)=0,\ y'(0)=\tfrac12$. (b) A second-order linear ODE in $x$ with no $y$ term, so it is reducible to first order.

Find. (a) The unique solution $y(t)$. (b) The general solution $y(x)$ (two arbitrary constants).

Approach. (a) Solve the homogeneous equation, recognize that the forcing frequency coincides with a homogeneous root (resonance) so the particular-solution trial needs an extra factor of $t$, then fix constants from the ICs. (b) Substitute $p=y'$ to drop the order by one, solve the resulting first-order linear ODE in $p$ by an integrating factor, then integrate twice.

(a) Homogeneous solution. Characteristic equation $r^2-r-6=0=(r-3)(r+2)$, so $r=3,-2$ and $$y_h(t)=C_1e^{3t}+C_2e^{-2t}.$$
(a) Particular solution — resonance case. The forcing $2e^{-2t}$ matches the homogeneous root $r=-2$, so the plain trial $Ae^{-2t}$ would solve the homogeneous equation; try instead $y_p=Ate^{-2t}$. Differentiating twice and substituting, $$y_p''-y_p'-6y_p = -5Ae^{-2t}.$$ Matching to $2e^{-2t}$ gives $A=-\tfrac25$, so $y_p(t)=-\tfrac25 te^{-2t}$.
(a) Apply the initial conditions. General solution $y(t)=C_1e^{3t}+C_2e^{-2t}-\tfrac25te^{-2t}$. $y(0)=C_1+C_2=0$. Differentiating, $y'(t)=3C_1e^{3t}-2C_2e^{-2t}-\tfrac25e^{-2t}+\tfrac45te^{-2t}$, so $y'(0)=3C_1-2C_2-\tfrac25=\tfrac12\Rightarrow 3C_1-2C_2=\tfrac9{10}$. With $C_2=-C_1$: $5C_1=\tfrac9{10}\Rightarrow C_1=\tfrac9{50},\ C_2=-\tfrac9{50}$. $$y(t)=\boxed{\tfrac{9}{50}e^{3t}-\tfrac{9}{50}e^{-2t}-\tfrac25te^{-2t}}$$
(b) Reduce the order. Let $p=y'$; then $xy''-y'=3x^2e^x$ becomes $xp'-p=3x^2e^x$, i.e. $p'-\tfrac1xp=3xe^x$ — a first-order linear ODE in $p$. Its integrating factor is $\mu=e^{-\int dx/x}=\tfrac1x$, giving $$\left(\frac{p}{x}\right)'=3e^x \;\Rightarrow\; \frac{p}{x}=3e^x+K_2 \;\Rightarrow\; p=3xe^x+K_2x.$$
(b) Integrate once more. $y'=p=3xe^x+K_2x$. Integrating (by parts on the first term, $\int xe^x\,dx=xe^x-e^x$), $$y(x)=3(xe^x-e^x)+\tfrac{K_2}{2}x^2+K_1.$$ Relabelling $\tfrac{K_2}{2}\to C_2$ and $K_1\to C_1$, $$\Rightarrow\quad y(x)=\boxed{C_1+C_2x^2+3xe^x-3e^x}$$

Part	Result
(a) $y(t)$	$\tfrac{9}{50}e^{3t}-\tfrac{9}{50}e^{-2t}-\tfrac25te^{-2t}$
(b) $y(x)$	$C_1+C_2x^2+3xe^x-3e^x$

Concept & Theory

Topic. Constant-coefficient linear ODEs with resonant forcing; order reduction for ODEs missing the dependent variable.

Key relations. Resonance trial: if the forcing $e^{kt}$ matches a simple homogeneous root $r=k$, multiply the usual trial by $t$ (by $t^2$ if $k$ is a double root). For an ODE $F(x,y',y'')=0$ with no explicit $y$, the substitution $p=y'$ reduces it to a first-order ODE in $p(x)$.

Why this works. A bare trial $Ae^{-2t}$ in (a) is itself a homogeneous solution, so it would produce $0=2e^{-2t}$ on substitution — inconsistent. Multiplying by $t$ breaks that degeneracy while keeping the same exponential envelope. In (b), $y$ never appears alone, so treating $p=y'$ as the new unknown converts a 2nd-order ODE into a 1st-order one, solvable by an integrating factor, before integrating once more to recover $y$.

Common pitfall. In (a), using the non-resonant trial $Ae^{-2t}$ and getting a contradictory equation instead of recognizing resonance. In (b), forgetting the integration constant $K_1$ that appears in the *second* integration (only $K_2$ shows up from solving for $p$) — the general solution needs exactly two independent constants for a 2nd-order ODE.

Source. Kreyszig, Advanced Engineering Mathematics, Ch. 2 (§2.7 Undetermined Coefficients — resonance) and Ch. 1 (reducible higher-order ODEs).

Question 1 of 8: Two Linear ODEs — Resonant IVP and a Reducible Second-Order Equation

Question 1: Two Linear ODEs — Resonant IVP and a Reducible Second-Order Equation (a) 10, (b) 10 marks