Second-order linear differential equations with constant coefficients

Worked solutions

Problem 1

1. $r^2 + r - 6 = (r + 3)(r - 2) = 0$, so $r = 2$ or $r = -3$Put $y = e^{rt}$ in and divide by $e^{rt}$, which is never $0$.
2. $y = c_1e^{2t} + c_2e^{-3t}$Two distinct real roots, two exponentials.
3. $y(0) = c_1 + c_2 = 1$ and $y'(0) = 2c_1 - 3c_2 = 7$Differentiate before substituting $t = 0$.
4. $2c_1 - 3(1 - c_1) = 7$, so $5c_1 = 10$ and $c_1 = 2$, $c_2 = -1$Substitute $c_2 = 1 - c_1$.
5. $y = 2e^{2t} - e^{-3t}$Check: $y'' + y' - 6y = (8 + 4 - 12)e^{2t} + (-9 + 3 + 6)e^{-3t} = 0$, and $y'(0) = 4 + 3 = 7$. The $e^{2t}$ term takes over for large $t$.

Problem 2

1. $r^2 + 6r + 9 = (r + 3)^2 = 0$, so $r = -3$ twiceThe discriminant $36 - 36$ is $0$.
2. $y = (c_1 + c_2t)e^{-3t}$A repeated root gives $e^{-3t}$ and $te^{-3t}$. Check the second: $(te^{-3t})'' + 6(te^{-3t})' + 9te^{-3t} = (9t - 6) + 6(1 - 3t) + 9t = 0$, times $e^{-3t}$.
3. $y(0) = c_1 = 2$$t = 0$ removes the $c_2$ term.
4. $y' = \big(c_2 - 3(c_1 + c_2t)\big)e^{-3t}$, so $y'(0) = c_2 - 3c_1 = -1$ and $c_2 = 5$The product rule on $(c_1 + c_2t)e^{-3t}$.
5. $y = (2 + 5t)e^{-3t}$Check: $y' = (-1 - 15t)e^{-3t}$, so $y'(0) = -1$, and $y'' + 6y' + 9y = 0$. One exponential alone could not meet both conditions (Mistake 1).

Problem 3

1. $r = \dfrac{-2 \pm \sqrt{4 - 20}}{2} = \dfrac{-2 \pm 4i}{2} = -1 \pm 2i$The discriminant is $-16$, so the roots are complex.
2. $y = e^{-t}(c_1\cos2t + c_2\sin2t)$$\alpha = -1$ is the real part and $\beta = 2$ the imaginary part.
3. $y(0) = c_1 = 1$$\cos0 = 1$ and $\sin0 = 0$.
4. $y' = e^{-t}\big((2c_2 - c_1)\cos2t - (2c_1 + c_2)\sin2t\big)$, so $y'(0) = 2c_2 - c_1 = 3$ and $c_2 = 2$The product rule, then $t = 0$.
5. $y = e^{-t}(\cos2t + 2\sin2t)$An oscillation with period $\tfrac{2\pi}{2} = \pi$ inside the decaying envelope $\sqrt5\,e^{-t}$, where $\sqrt{1^2 + 2^2} = \sqrt5$.

Problem 4

1. $x'' + 25x = 0$, so $r^2 + 25 = 0$ and $r = \pm5i$Divide by the mass. $\omega = \sqrt{k/m} = \sqrt{50/2} = 5$ rad/s.
2. $x = A\cos5t + B\sin5t$$\alpha = 0$: no decay.
3. $x(0) = A = 0.1$ and $x'(0) = 5B = 1$, so $B = 0.2$$x' = -5A\sin5t + 5B\cos5t$.
4. Amplitude $\sqrt{0.1^2 + 0.2^2} = \sqrt{0.05} \approx 0.2236$ m$A\cos\omega t + B\sin\omega t = R\cos(\omega t - \varphi)$ with $R = \sqrt{A^2 + B^2}$.
5. $x = 0.1\cos5t + 0.2\sin5t$; amplitude $\approx 0.224$ m, period $\tfrac{2\pi}{5} \approx 1.257$ s, frequency $\tfrac{5}{2\pi} \approx 0.796$ cycles per secondCheck: $x\big(\tfrac{2\pi}{5}\big) = 0.1$, back where it started. The period depends only on $m$ and $k$, not on how far the spring is pulled.

Problem 5

1. $r^2 + cr + 4 = 0$ has discriminant $c^2 - 16$Critical damping is where it is $0$: $c = 2\sqrt{mk} = 4$.
2. $c = 1$: $r = \dfrac{-1 \pm i\sqrt{15}}{2}$, underdampedIt oscillates at $\beta = \tfrac{\sqrt{15}}{2} \approx 1.936$ rad/s, a little slower than the undamped $2$, inside the envelope $e^{-t/2}$.
3. $c = 5$: $r^2 + 5r + 4 = (r + 1)(r + 4)$, so $r = -1, -4$, overdampedNo oscillation; the slow mode $e^{-t}$ sets how fast it settles.
4. $c = 4$: $r = -2$ twice, so $x = (c_1 + c_2t)e^{-2t}$ with $c_1 = 1$ and $c_2 - 2c_1 = 0$$x'(0) = c_2 - 2c_1$, as in Problem 2.
5. Underdamped for $c = 1$, critically damped for $c = 4$, overdamped for $c = 5$; $x = (1 + 2t)e^{-2t}$ at $c = 4$The critically damped spring settles fastest without overshooting, as $e^{-2t}$: more damping ($e^{-t}$ for $c = 5$) or less ($e^{-t/2}$ for $c = 1$) is slower.

Problem 6

1. $r^2 + r - 2 = (r + 2)(r - 1)$, so $y_h = c_1e^t + c_2e^{-2t}$The homogeneous equation first.
2. Guess $y_p = At^2 + Bt + C$, so $y_p' = 2At + B$ and $y_p'' = 2A$A quadratic input needs a full quadratic guess, lower terms included.
3. $2A + (2At + B) - 2(At^2 + Bt + C) = -2At^2 + (2A - 2B)t + (2A + B - 2C) = 4t^2$Collect powers of $t$.
4. $-2A = 4$, $2A - 2B = 0$, $2A + B - 2C = 0$ give $A = -2$, $B = -2$, $C = -3$Match the coefficients of $t^2$, $t$ and $1$.
5. $y = \tfrac83e^t + \tfrac13e^{-2t} - 2t^2 - 2t - 3$The conditions go on the full solution: $c_1 + c_2 - 3 = 0$ and $c_1 - 2c_2 - 2 = 0$, so $c_1 = \tfrac83$, $c_2 = \tfrac13$. Check: $y(0) = \tfrac83 + \tfrac13 - 3 = 0$.

Problem 7

1. $r^2 - r - 6 = (r - 3)(r + 2)$, so $y_h = c_1e^{3t} + c_2e^{-2t}$Find the homogeneous solutions before guessing.
2. $y_p = Ae^t$: $A(1 - 1 - 6)e^t = -6Ae^t = 12e^t$, so $A = -2$$e^t$ is not a homogeneous solution, so the plain guess works.
3. $Ae^{3t}$ fails for the second input: it gives $A(9 - 3 - 6)e^{3t} = 0$$e^{3t}$ already solves the homogeneous equation (Mistake 3).
4. $y_p = Ate^{3t}$: $y_p' = A(1 + 3t)e^{3t}$, $y_p'' = A(6 + 9t)e^{3t}$, and $A\big((6 + 9t) - (1 + 3t) - 6t\big)e^{3t} = 5Ae^{3t} = 10e^{3t}$The $t$ terms cancel, as they must; $A = 2$.
5. $y_p = -2e^t$ for $12e^t$, and $y_p = 2te^{3t}$ for $10e^{3t}$The general solution of the second is $c_1e^{3t} + c_2e^{-2t} + 2te^{3t}$.

Problem 8

1. $x_p = A\cos t + B\sin t$, $x_p' = -A\sin t + B\cos t$, $x_p'' = -A\cos t - B\sin t$A cosine input needs both a cosine and a sine in the guess.
2. Cosines: $-A + 2B + 5A = 4A + 2B = 10$; sines: $-B - 2A + 5B = 4B - 2A = 0$Collect $\cos t$ and $\sin t$ separately.
3. $A = 2B$, so $10B = 10$: $B = 1$ and $A = 2$The homogeneous part $e^{-t}(c_1\cos2t + c_2\sin2t)$ dies out, so $x_p$ is what remains.
4. Amplitude $\sqrt{2^2 + 1^2} = \sqrt5 \approx 2.236$, lag $\varphi = \arctan\tfrac12 \approx 0.464$ rad$2\cos t + \sin t = \sqrt5\cos(t - \varphi)$. The same comes from $\dfrac{10}{\sqrt{(5 - \omega^2)^2 + (2\omega)^2}} = \dfrac{10}{\sqrt{20}}$ at $\omega = 1$.
5. $x_p = 2\cos t + \sin t$, amplitude $\sqrt5$, lag $\approx 0.464$ rad; the amplitude is largest at $\omega = \sqrt3$, where it is $2.5$Minimize $(5 - \omega^2)^2 + 4\omega^2 = \omega^4 - 6\omega^2 + 25$: its derivative in $\omega^2$ is $2\omega^2 - 6$, zero at $\omega^2 = 3$, where it equals $16$ and the amplitude is $\tfrac{10}{4}$. Damping moves the peak below the natural frequency $\sqrt5$ and keeps it finite.

Problem 9

1. $r^2 + 4 = 0$, so $y_h = c_1\cos2t + c_2\sin2t$$r = \pm2i$.
2. For $8t$: $y_1 = at + b$ gives $4at + 4b = 8t$, so $y_1 = 2t$$y_1'' = 0$.
3. For $5e^{-t}$: $y_2 = Ce^{-t}$ gives $(1 + 4)Ce^{-t} = 5e^{-t}$, so $y_2 = e^{-t}$Superposition: solve for each input separately and add.
4. $y(0) = c_1 + 1 = 1$ and $y'(0) = 2c_2 + 2 - 1 = 0$, so $c_1 = 0$ and $c_2 = -\tfrac12$$y' = -2c_1\sin2t + 2c_2\cos2t + 2 - e^{-t}$.
5. $y = 2t + e^{-t} - \tfrac12\sin2t$Check: $y'' = e^{-t} + 2\sin2t$ and $4y = 8t + 4e^{-t} - 2\sin2t$ add to $8t + 5e^{-t}$.

Problem 10

1. $A = \begin{pmatrix}0 & 1\\ -2 & -3\end{pmatrix}$The first row says $(y)' = y'$; the second says $(y')' = -2y - 3y'$.
2. $\det(A - rI) = (-r)(-3 - r) + 2 = r^2 + 3r + 2$The characteristic polynomial of $A$ is the characteristic polynomial of the equation, so the eigenvalues are $r = -1$ and $r = -2$.
3. Eigenvectors $(1, -1)$ for $r = -1$ and $(1, -2)$ for $r = -2$The first row of $(A - rI)\mathbf v = 0$ is $-rv_1 + v_2 = 0$, so $\mathbf v = (1, r)$: a solution $e^{rt}$ and its derivative $re^{rt}$.
4. $\mathbf x = c_1e^{-t}(1, -1) + c_2e^{-2t}(1, -2)$ with $c_1 + c_2 = 1$ and $-c_1 - 2c_2 = 0$So $c_1 = 2$ and $c_2 = -1$.
5. $y = 2e^{-t} - e^{-2t}$Every equation here is a $2\times2$ system in disguise. Gradient descent with momentum on a quadratic of curvature $\lambda$ behaves like $y'' + \gamma y' + \lambda y = 0$, and it converges fastest at the critical damping $\gamma = 2\sqrt\lambda$, as in Problem 5.