Math 308 Week in Review 5 (Exam 1 Review) — Solutions

(a) $y'' + 3y' + 2y = 5e^{4t}$

Characteristic equation: $r^2 + 3r + 2 = (r+1)(r+2) = 0$, so $r = -1, -2$.

Homogeneous solution: $y_c = c_1 e^{-t} + c_2 e^{-2t}$.

The forcing is $5e^{4t}$. Since $r = 4$ is not a root of the characteristic equation, there is no overlap.

(b) $y'' + y' - 6y = 2t^2 + 1$

Characteristic equation: $r^2 + r - 6 = (r+3)(r-2) = 0$, so $r = -3, 2$.

Homogeneous solution: $y_c = c_1 e^{-3t} + c_2 e^{2t}$.

The forcing is a polynomial of degree 2. Since $r = 0$ is not a root, there is no overlap.

(c) $y'' + 4y = 3\cos(5t)$

Characteristic equation: $r^2 + 4 = 0$, so $r = \pm 2i$.

Homogeneous solution: $y_c = c_1 \cos(2t) + c_2 \sin(2t)$.

The forcing involves $\cos(5t)$. Since $\pm 5i$ are not roots, there is no overlap.

(d) $y'' - 5y' + 6y = 3e^{2t}$

Characteristic equation: $r^2 - 5r + 6 = (r-2)(r-3) = 0$, so $r = 2, 3$.

Homogeneous solution: $y_c = c_1 e^{2t} + c_2 e^{3t}$.

The forcing is $3e^{2t}$. Since $r = 2$ is a simple root, $Ae^{2t}$ duplicates $y_c$. Multiply by $t$ ($s = 1$).

(e) $y'' - 4y' + 4y = 5e^{2t}$

Characteristic equation: $r^2 - 4r + 4 = (r-2)^2 = 0$, so $r = 2$ (repeated).

Homogeneous solution: $y_c = c_1 e^{2t} + c_2\,te^{2t}$.

The forcing is $5e^{2t}$. Since $r = 2$ is a repeated root (multiplicity 2), both $Ae^{2t}$ and $Ate^{2t}$ duplicate $y_c$. Multiply by $t^2$ ($s = 2$).

(f) $y'' + 9y = \cos(3t)$

Characteristic equation: $r^2 + 9 = 0$, so $r = \pm 3i$.

Homogeneous solution: $y_c = c_1 \cos(3t) + c_2 \sin(3t)$.

The forcing involves $\cos(3t)$. Since $\pm 3i$ are roots, $A\cos(3t) + B\sin(3t)$ duplicates $y_c$. Multiply by $t$ ($s = 1$).

(g) $y'' + y' - 2y = 3t^2 - 1 + 4e^{-2t}$

Characteristic equation: $r^2 + r - 2 = (r+2)(r-1) = 0$, so $r = -2, 1$.

Homogeneous solution: $y_c = c_1 e^{-2t} + c_2 e^{t}$.

Split the forcing into two parts and apply superposition:

• For $3t^2 - 1$ (polynomial, degree 2): $r = 0$ is not a root, so use $At^2 + Bt + C$.

• For $4e^{-2t}$: $r = -2$ is a simple root, so $De^{-2t}$ duplicates $y_c$. Multiply by $t$ ($s = 1$): $Dte^{-2t}$.

(h) $y'' - y = te^{t}$

Characteristic equation: $r^2 - 1 = (r-1)(r+1) = 0$, so $r = 1, -1$.

Homogeneous solution: $y_c = c_1 e^{t} + c_2 e^{-t}$.

The forcing is $te^t$. Without duplication we would try $(At + B)e^t$. But $r = 1$ is a simple root, so $e^t$ terms duplicate $y_c$. Multiply the entire trial by $t$ ($s = 1$):

(i) $y'' + 4y = t\sin(2t)$

Characteristic equation: $r^2 + 4 = 0$, so $r = \pm 2i$.

Homogeneous solution: $y_c = c_1 \cos(2t) + c_2 \sin(2t)$.

The forcing is $t\sin(2t)$. Without duplication we would try $(At + B)\cos(2t) + (Ct + D)\sin(2t)$. But $r = \pm 2i$ are roots, so $\cos(2t)$ and $\sin(2t)$ terms duplicate $y_c$. Multiply by $t$ ($s = 1$):

The characteristic equation is $r^2 + 3r + 2 = (r+1)(r+2) = 0$, so $r = -1, -2$.

Since $g(t) = 6t + 4$ is a first-degree polynomial and $r = 0$ is not a root, try $y_p(t) = At + B$.

Then $y_p' = A$ and $y_p'' = 0$. Substituting into the ODE:

Matching coefficients:

$t^1$: $2A = 6 \implies A = 3$

$t^0$: $3A + 2B = 4 \implies 9 + 2B = 4 \implies B = -\dfrac{5}{2}$

So $y_p(t) = 3t - \dfrac{5}{2}$.

The general solution is $y = y_c + y_p$:

The characteristic equation is $r^2 - 2r - 3 = (r-3)(r+1) = 0$, so $r = 3, -1$.

The forcing is $5e^{3t}$. Since $r = 3$ is a simple root of the characteristic equation, we must multiply by $t$. Try $y_p(t) = Ate^{3t}$.

$y_p' = Ae^{3t} + 3Ate^{3t} = A(1 + 3t)e^{3t}$

$y_p'' = A(6 + 9t)e^{3t}$

Substituting into $y'' - 2y' - 3y = 5e^{3t}$:

Collecting terms: $A[6 + 9t - 2 - 6t - 3t]e^{3t} = A \cdot 4 \cdot e^{3t} = 5e^{3t}$.

So $4A = 5$, giving $A = \dfrac{5}{4}$.

The general solution is:

Applying initial conditions:

$y(0) = c_1 + c_2 = 1$

$y'(t) = 3c_1 e^{3t} - c_2 e^{-t} + \dfrac{5}{4}e^{3t} + \dfrac{15}{4}\,te^{3t}$

$y'(0) = 3c_1 - c_2 + \dfrac{5}{4} = 0 \implies 3c_1 - c_2 = -\dfrac{5}{4}$

Adding the two equations: $4c_1 = 1 - \dfrac{5}{4} = -\dfrac{1}{4}$, so $c_1 = -\dfrac{1}{16}$.

Then $c_2 = 1 - c_1 = 1 + \dfrac{1}{16} = \dfrac{17}{16}$.

Divide through by $t^2$ to put in standard form:

So $g(t) = 3 - t^{-2}$.

The Wronskian of $y_1 = t^2$, $y_2 = t^{-1}$ is:

$y_1' = 2t$, $\quad y_2' = -t^{-2}$.

Using the variation of parameters formulas:

Integrating:

The particular solution is $y_p = u_1 y_1 + u_2 y_2$:

The $-t^2/3$ term is a multiple of $y_1 = t^2$ and the constant $1/2$ can be absorbed into the general solution (since we only need a particular solution), but we can keep them for completeness. A valid particular solution is:

(a) $ty' + y = t^2$

Order: 1 (highest derivative is $y'$). The dependent variable $y$ and its derivative $y'$ appear only to the first power and are not multiplied together. The coefficient $t$ is a function of the independent variable only. Linear.

(b) $y'' + yy' = 0$

Order: 2. The term $yy'$ is the product of $y$ and $y'$, which is nonlinear in the dependent variable. Nonlinear.

(c) $y''' - (\sin t)\,y = e^t$

Order: 3. Both $y'''$ and $y$ appear to the first power; $\sin t$ is a coefficient depending only on $t$. Linear.

(d) $(y')^2 + y = 1$

Order: 1. The term $(y')^2$ means $y'$ is raised to the second power, which is nonlinear. Nonlinear.

(a) Set $f(y) = y(y-2)(y-5) = 0$. The equilibrium solutions are:

(b) Check the sign of $f(y)$ in each interval:

• $y < 0$: e.g. $f(-1) = (-1)(-3)(-6) = -18 < 0$ — solutions decrease $(\leftarrow)$

• $0 < y < 2$: e.g. $f(1) = (1)(-1)(-4) = 4 > 0$ — solutions increase $(\rightarrow)$

• $2 < y < 5$: e.g. $f(3) = (3)(1)(-2) = -6 < 0$ — solutions decrease $(\leftarrow)$

• $y > 5$: e.g. $f(6) = (6)(4)(1) = 24 > 0$ — solutions increase $(\rightarrow)$

Phase line:

$y = 0$: unstable; $\quad$ $y = 2$: stable; $\quad$ $y = 5$: unstable

(c) Since $y(0) = 3$ lies in the interval $2 < y < 5$ where $f(y) < 0$, solutions decrease. The solution is bounded below by the stable equilibrium $y = 2$.

Characteristic equation: $r^2 - 2r + 5 = 0$.

General solution:

Since the real part of the roots is $\lambda = 1 > 0$, the factor $e^{t} \to \infty$ as $t \to \infty$. The solutions exhibit growing oscillations: they oscillate with exponentially increasing amplitude.

Separate variables (assuming $y \neq 0$):

Integrate both sides:

Apply the initial condition $y(1) = 1$:

So $-\dfrac{1}{y} = x^2 - 2$, which gives $y = \dfrac{1}{2 - x^2}$.

We can verify: $y(1) = \dfrac{1}{2 - 1} = 1$ $\checkmark$

Divide by $t^2$ to get standard form:

So $p(t) = -2/t$.

Set $y = u(t) \cdot t$. Then $y' = u't + u$ and $y'' = u''t + 2u'$. Substituting into the standard form:

Since $t > 0$, we have $u'' = 0$.

Integrating twice: $u'' = 0 \implies u' = A \implies u = At + B$. Taking $A = 1$, $B = 0$: $u = t$.

Therefore the second solution is:

The equation is already in standard form $y' + p(t)y = g(t)$ with $p(t) = 2$.

The integrating factor is:

Multiplying both sides by $\mu$:

The left side is $\dfrac{d}{dt}\big[e^{2t}y\big]$.

Integrating both sides:

Solving for $y$:

(a) If $y(t) = 1$ for all $t$, then $y' = 0$ and $\sqrt{y - 1} = \sqrt{0} = 0$. So $y' = \sqrt{y-1}$ is satisfied. Also, $y(2) = 1$ $\checkmark$. Therefore $y(t) = 1$ is indeed a solution.

(b) The theorem (Theorem 2.4.2) guarantees a unique solution if both $f(t, y) = \sqrt{y-1}$ and $\dfrac{\partial f}{\partial y} = \dfrac{1}{2\sqrt{y-1}}$ are continuous in a rectangle containing the initial point $(2, 1)$.

At $y = 1$: $\dfrac{\partial f}{\partial y} = \dfrac{1}{2\sqrt{1-1}} = \dfrac{1}{0}$, which is undefined.

Since $\partial f / \partial y$ is not continuous at $y = 1$, the hypotheses of the theorem are not satisfied. The theorem does not guarantee uniqueness.

In fact, uniqueness fails: one can verify that $y(t) = 1 + \dfrac{1}{4}(t-2)^2$ for $t \geq 2$ (and $y = 1$ for $t < 2$) is another solution through $(2, 1)$.

(a) $r = 4$ and $r = -1$ (distinct real roots):

(b) $r = -3$ (repeated):

(c) $r = 1 \pm 2i$ (complex conjugate, $\lambda = 1$, $\mu = 2$):

Let $M(x,y) = 2xy + 3$ and $N(x,y) = x^2 + 4y$. Check exactness:

Since $M_y = N_x = 2x$, the equation is exact.

We need $\psi(x,y)$ such that $\psi_x = M$ and $\psi_y = N$. Integrating $\psi_x = 2xy + 3$ with respect to $x$:

Differentiating with respect to $y$ and setting equal to $N$:

So $h'(y) = 4y$, giving $h(y) = 2y^2$.

The implicit solution is $\psi(x,y) = C$:

(a) Set up the IVP.

Rate in: $(2\;\text{g/L})(3\;\text{L/min}) = 6$ g/min.

Rate out: $\left(\dfrac{Q(t)}{100}\;\text{g/L}\right)(3\;\text{L/min}) = \dfrac{3Q}{100}$ g/min.

(Note: the volume stays constant at 100 L since inflow rate = outflow rate.)

(b) Solve.

Rewrite as $Q' + \dfrac{3}{100}\,Q = 6$. This is first-order linear.

Integrating factor: $\mu = e^{3t/100}$.

Apply $Q(0) = 0$: $0 = 200 + C$, so $C = -200$.

(c) Limiting value.

As $t \to \infty$, $e^{-3t/100} \to 0$, so:

(a) The slopes are zero when $y' = y(1-y) = 0$, i.e., at $y = 0$ and $y = 1$.

Sign analysis of $f(y) = y(1-y)$:

• $y < 0$: $y < 0$ and $1-y > 0$, so $f(y) < 0$. Slopes are negative.

• $0 < y < 1$: $y > 0$ and $1-y > 0$, so $f(y) > 0$. Slopes are positive.

• $y > 1$: $y > 0$ and $1-y < 0$, so $f(y) < 0$. Slopes are negative.

Direction field:

(b) $y(0) = 0.5$ is in the region $0 < y < 1$ where $f(y) > 0$, so the solution increases. It is bounded above by the equilibrium $y = 1$ (which is stable, since $f > 0$ below and $f < 0$ above).

(c) $y(0) = 1.5$ is in the region $y > 1$ where $f(y) < 0$, so the solution decreases. It approaches the equilibrium $y = 1$ from above.

(a) If $W[y_1, y_2](t_0) \neq 0$ at some $t_0$ in $I$, then $y_1$ and $y_2$ are linearly independent on $I$ and form a fundamental set of solutions. Every solution of the ODE can be written as $c_1 y_1 + c_2 y_2$.

(b) If $W[y_1, y_2](t_0) = 0$ at some $t_0$ in $I$, then by Abel's Theorem, $W(t) = 0$ for all $t$ in $I$. This means $y_1$ and $y_2$ are linearly dependent on $I$. They do not form a fundamental set of solutions.

(c) False. By Abel's Theorem, for solutions of $y'' + p(t)y' + q(t)y = 0$ (with $p, q$ continuous), the Wronskian satisfies $W(t) = W(t_0)\,e^{-\int p(t)\,dt}$. Since the exponential function is never zero, $W(t)$ is either always zero (if $W(t_0) = 0$) or never zero (if $W(t_0) \neq 0$) on $I$. It cannot be zero at one point and nonzero at another.

The general solution involves $e^{-t}$ and $te^{-t}$, which tells us $r = -1$ is a repeated root.

The characteristic equation is:

Therefore the differential equation is:

Verification:

$y = Ce^{-2t} + 1 \implies y' = -2Ce^{-2t}$.

Substituting into $y' + 2y$:

So $y' + 2y = 2$ is satisfied for every value of $C$.

Particular solution with $y(0) = 4$:

$y(0) = C \cdot 1 + 1 = C + 1 = 4$, so $C = 3$.

Dividing by $t-1$ to get standard form:

So $p(t) = \dfrac{t}{t-1}$ and $q(t) = \dfrac{-2}{t-1}$.

Both $p(t)$ and $q(t)$ are discontinuous at $t = 1$ (where $t - 1 = 0$). They are continuous everywhere else.

The discontinuity at $t = 1$ divides the real line into two intervals: $(-\infty, 1)$ and $(1, \infty)$. Since the initial point $t_0 = 2$ lies in the interval $(1, \infty)$, the theorem guarantees a unique solution on this interval.

The characteristic equation is $r^2 + 1 = 0$, giving $r = \pm i$.

So $y_1 = \cos t$, $y_2 = \sin t$, and $g(t) = \sec t$.

The Wronskian is:

Using the variation of parameters formulas:

Integrating:

So $y_p = u_1 y_1 + u_2 y_2 = \cos t \cdot \ln|\cos t| + t\sin t$, and the general solution is: