Directions: Full worked solutions for Week in Review 4 (Differential Equations). Click Solution under any problem to reveal the worked answer. Use the Contents panel or the Prev / Up / Next buttons to move between sections, and Search to find terms on this page.
Solve the initial value problem $y'' + 3y' + 2y = 0$, $y(0) = 0$, $y'(0) = -1$.
The characteristic equation is $r^2 + 3r + 2 = (r+1)(r+2) = 0$, giving $r = -1, -2$.
General solution:
Applying initial conditions:
$y(0) = c_1 + c_2 = 0$
$y'(0) = -c_1 - 2c_2 = -1$
From the first equation: $c_1 = -c_2$. Substituting into the second: $c_2 - 2c_2 = -c_2 = -1$, so $c_2 = 1$ and $c_1 = -1$.
Solve the initial value problem $y'' - 6y' + 9y = 0$, $y(0) = 2$, $y'(0) = -1$.
The characteristic equation is $r^2 - 6r + 9 = (r - 3)^2 = 0$, giving repeated root $r = 3$.
General solution:
Applying initial conditions:
$y(0) = c_1 = 2$
$y'(t) = 3c_1 e^{3t} + c_2 e^{3t} + 3c_2 te^{3t}$, so $y'(0) = 3c_1 + c_2 = 6 + c_2 = -1$, giving $c_2 = -7$.
Solve the initial value problem $y'' - 4y' + 13y = 0$, $y(0) = 0$, $y'(0) = 3$.
The characteristic equation is $r^2 - 4r + 13 = 0$:
General solution:
$y(0) = c_1 = 0$.
$y'(t) = 2e^{2t}(c_1\cos 3t + c_2\sin 3t) + e^{2t}(-3c_1\sin 3t + 3c_2\cos 3t)$, so $y'(0) = 2c_1 + 3c_2 = 3c_2 = 3$, giving $c_2 = 1$.
Find the initial value problem (the differential equation and the initial conditions) that has the solution $y(t) = 3e^{2t} - e^{5t}$.
The solution involves $e^{2t}$ and $e^{5t}$, so the characteristic roots are $r = 2$ and $r = 5$.
Characteristic equation: $(r - 2)(r - 5) = r^2 - 7r + 10 = 0$.
The differential equation is $y'' - 7y' + 10y = 0$.
Initial conditions: $y(0) = 3 - 1 = 2$.
$y'(t) = 6e^{2t} - 5e^{5t}$, so $y'(0) = 6 - 5 = 1$.
Find the initial value problem (the differential equation and the initial conditions) that has the solution $y(t) = 3e^{-t} - 5te^{-t}$.
The solution involves $e^{-t}$ and $te^{-t}$, so $r = -1$ is a repeated root.
Characteristic equation: $(r + 1)^2 = r^2 + 2r + 1 = 0$.
The differential equation is $y'' + 2y' + y = 0$.
Initial conditions: $y(0) = 3$.
$y'(t) = -3e^{-t} - 5e^{-t} + 5te^{-t} = (-8 + 5t)e^{-t}$, so $y'(0) = -8$.
Find the initial value problem (the differential equation and the initial conditions) that has the solution $y(t) = e^{3t}\cos(4t) - 2e^{3t}\sin(4t)$.
The solution has the form $e^{3t}(c_1\cos 4t + c_2\sin 4t)$, so the roots are $r = 3 \pm 4i$.
Characteristic equation: $(r-3)^2 + 16 = r^2 - 6r + 25 = 0$.
The differential equation is $y'' - 6y' + 25y = 0$.
Initial conditions: $y(0) = 1\cdot\cos 0 - 2\cdot\sin 0 = 1$.
$y'(t) = 3e^{3t}(\cos 4t - 2\sin 4t) + e^{3t}(-4\sin 4t - 8\cos 4t)$
$y'(t) = e^{3t}[(-5)\cos 4t + (-10)\sin 4t]$
$y'(0) = -5$.
Verify that $y_1(t) = t^{-1}$ and $y_2(t) = t^{-1}\ln t$ are solutions of $t^2y'' + 3ty' + y = 0$, $t > 0$. Do they constitute a fundamental set?
Verify $y_1 = t^{-1}$: $y_1' = -t^{-2}$, $y_1'' = 2t^{-3}$.
Verify $y_2 = t^{-1}\ln t$: $y_2' = t^{-2}(1 - \ln t)$, $y_2'' = t^{-3}(2\ln t - 3)$.
Wronskian:
Since $W(t) = t^{-3} \neq 0$ for all $t > 0$, they form a fundamental set.
Suppose $y_1(t) = t^{-2}$ is a solution of $t^2y'' + 5ty' + 4y = 0$, $t > 0$. Find a second linearly independent solution.
Put in standard form: $y'' + \dfrac{5}{t}\,y' + \dfrac{4}{t^2}\,y = 0$, so $p(t) = 5/t$.
Set $y = u(t)\cdot t^{-2}$. Substituting and simplifying (the $u$-terms cancel), we obtain:
Let $w = u'$: $w' + \dfrac{1}{t}\,w = 0$. Separating: $\dfrac{dw}{w} = -\dfrac{dt}{t}$, so $w = A/t$.
Integrating: $u = A\ln t$. Taking $A = 1$:
Given that $y_1(t) = e^{t}$ is a solution of $ty'' - (t+1)y' + y = 0$, $t > 0$, use reduction of order to find a second linearly independent solution.
Standard form: $y'' - \dfrac{t+1}{t}\,y' + \dfrac{1}{t}\,y = 0$.
Set $y = u(t)\,e^{t}$. Then $y' = (u' + u)e^t$ and $y'' = (u'' + 2u' + u)e^t$. Substituting and dividing by $e^t$:
Expanding: $tu'' + 2tu' + tu - tu' - tu - u' - u + u = 0$, which simplifies to:
Let $w = u'$: $tw' + (t-1)w = 0$, i.e., $w' = \left(\dfrac{1}{t} - 1\right)w$.
Separating: $\ln|w| = \ln t - t + C$, so $w = Ate^{-t}$.
Integrating by parts: $u = A\displaystyle\int te^{-t}\,dt = A[-(t+1)e^{-t}]$.
Taking $A = -1$: $u = (t+1)e^{-t}$, so:
Given that $y_1(t) = t\sin t$ is a solution of $t^2y'' - 2ty' + (t^2 + 2)y = 0$, $t > 0$, use reduction of order to find a second linearly independent solution and write the general solution.
Set $y = u(t)\cdot t\sin t$. After substituting into the ODE and simplifying (e.g., using SymPy), the terms involving $u$ cancel and we obtain:
Since $t > 0$, we need $\sin(t)\,u'' + 2\cos(t)\,u' = 0$.
Let $w = u'$: $\sin(t)\,w' + 2\cos(t)\,w = 0$. Separating:
Integrating: $u = A\displaystyle\int \csc^2 t\,dt = -A\cot t$. Taking $A = -1$: $u = \cot t$.
Therefore $y_2 = \cot t \cdot t\sin t = t\cos t$.
Compute the Wronskian for each pair and determine whether they form a fundamental set of solutions for the given equation.
(a) $y_1 = e^{2t}$, $y_2 = te^{2t}$, for $y'' - 4y' + 4y = 0$.
$y_1' = 2e^{2t}$, $y_2' = (1 + 2t)e^{2t}$.
$W = e^{4t} \neq 0$ for all $t$. The characteristic equation $(r-2)^2 = 0$ confirms $y_1, y_2$ are solutions. Yes, fundamental set.
(b) $y_1 = e^{3t}$, $y_2 = e^{-3t}$, for $y'' - 9y = 0$.
$y_1' = 3e^{3t}$, $y_2' = -3e^{-3t}$.
$W = -6 \neq 0$. The characteristic equation $r^2 - 9 = (r-3)(r+3) = 0$ confirms these are solutions. Yes, fundamental set.
Find the general solution of $2y'' + y' - y = 0$.
Characteristic equation: $2r^2 + r - 1 = (2r - 1)(r + 1) = 0$, giving $r = \dfrac{1}{2}$ and $r = -1$.
Write the general solution given the characteristic roots: (a) $r = -2 \pm 5i$; (b) $r = 3$ (repeated); (c) $r = 0$ and $r = -7$.
(a) Complex roots $r = -2 \pm 5i$ (here $\lambda = -2$, $\mu = 5$):
(b) Repeated real root $r = 3$:
(c) Distinct real roots $r = 0$ and $r = -7$:
Find the general solution of $y'' + 2y' + 5y = 0$ and describe the long-term behavior as $t \to \infty$.
Characteristic equation: $r^2 + 2r + 5 = 0$.
General solution:
The factor $e^{-t} \to 0$ as $t \to \infty$, while $\cos 2t$ and $\sin 2t$ remain bounded. Therefore all solutions exhibit decaying oscillations and approach zero.
Solve the initial value problem $y'' + 9y = 0$, $y(0) = 1$, $y'(0) = -6$. Describe the behavior as $t \to \infty$.
Characteristic equation: $r^2 + 9 = 0$, giving $r = \pm 3i$ (purely imaginary roots).
General solution: $y = c_1\cos 3t + c_2\sin 3t$.
$y(0) = c_1 = 1$.
$y'(t) = -3c_1\sin 3t + 3c_2\cos 3t$, so $y'(0) = 3c_2 = -6$, giving $c_2 = -2$.
The solution is purely oscillatory: there is no exponential factor, so it neither grows nor decays. The solution oscillates forever with constant amplitude.
Solve the IVP $y'' + 6y' + 25y = 0$, $y(0) = 2$, $y'(0) = -2$ and express the solution in amplitude-phase form $y(t) = Re^{\lambda t}\cos(\mu t - \delta)$.
Characteristic equation: $r^2 + 6r + 25 = 0$, giving $r = -3 \pm 4i$.
General solution: $y = e^{-3t}(c_1\cos 4t + c_2\sin 4t)$.
$y(0) = c_1 = 2$.
$y'(0) = -3c_1 + 4c_2 = -6 + 4c_2 = -2$, so $c_2 = 1$.
Amplitude-phase form: We write $c_1\cos\theta + c_2\sin\theta = R\cos(\theta - \delta)$ where $R = \sqrt{c_1^2 + c_2^2}$ and $\tan\delta = c_2/c_1$.
Let $y_1, y_2$ solve $y'' + p(t)y' + q(t)y = 0$ with $p, q$ continuous. (a) Compute $W(0)$ given $y_1(0)=1$, $y_1'(0)=0$, $y_2(0)=0$, $y_2'(0)=1$. (b) Can $y_1$ and $y_2$ both be zero at the same $t_0$? (c) If $W(t_0)=0$ for some $t_0$, what follows?
(a)
(b) Suppose $y_1(t_0) = y_2(t_0) = 0$ for some $t_0$. Then:
By Abel's Theorem, $W$ is either identically zero or never zero on any interval where $p$ and $q$ are continuous. If $y_1, y_2$ form a fundamental set, then $W \neq 0$ everywhere. So no, two members of a fundamental set cannot both be zero at the same point.
(c) If $W(t_0) = 0$ for some $t_0$ in the interval, then by Abel's Theorem $W(t) = 0$ for all $t$. This means $y_1$ and $y_2$ are linearly dependent, so they do not form a fundamental set.
For each equation, find the characteristic roots and describe the long-term behavior of solutions as $t \to \infty$.
(a) $y'' - y' - 6y = 0$: $r^2 - r - 6 = (r-3)(r+2) = 0$, so $r = 3, -2$.
Since $r = 3 > 0$, generic solutions grow exponentially.
(b) $y'' + 4y = 0$: $r^2 + 4 = 0$, so $r = \pm 2i$.
Pure imaginary roots: solutions oscillate with constant amplitude (neither grow nor decay).
(c) $y'' + 3y' = 0$: $r^2 + 3r = r(r+3) = 0$, so $r = 0, -3$.
$y = c_1 + c_2 e^{-3t}$. Solutions decay to a constant as $t \to \infty$.
(d) $y'' - 2y' + 10y = 0$: $r = 1 \pm 3i$.
Positive real part ($\lambda = 1$): solutions exhibit growing oscillations.
The equation $t^2y'' - 3ty' + 4y = 0$, $t > 0$ has a fundamental set with $W(2) = 8$. Find $W(3)$ without solving.
Dividing by $t^2$ to get standard form: $y'' - \dfrac{3}{t}\,y' + \dfrac{4}{t^2}\,y = 0$, so $p(t) = -3/t$.
By Abel's Theorem:
Computing the integral with $t_0 = 2$:
Therefore:
Evaluating at $t = 3$:
Suppose $y_1, y_2$ solve $y'' + 4y' - 5y = 0$ and $W(0) = 6$. Find $W(t)$ for all $t$.
The equation is already in standard form with $p(t) = 4$ (constant). By Abel's Theorem:
We can verify: the characteristic equation $r^2 + 4r - 5 = (r+5)(r-1) = 0$ gives $r = 1, -5$. With $y_1 = e^t, y_2 = e^{-5t}$:
This confirms the exponential form (the sign depends on the ordering of $y_1, y_2$).
