Math 308 Week in Review 3 (Solutions)

Here $f(t,y) = t^2 \cos y + e^{t}$. Computing the partial derivative:

The function $f$ is built from polynomials, trig functions, and exponentials—all continuous everywhere. Similarly, $f_y = -t^2 \sin y$ is continuous everywhere. Since the initial point $(0, 1)$ lies in a region where both $f$ and $f_y$ are continuous, the hypotheses are satisfied.

We have $M(t,y) = 2ty + 3$ and $N(t,y) = t^2 + 4y$. Checking exactness:

Since $M_y = N_t$, the equation is exact. Integrating $M$ with respect to $t$:

So we have:

To find $h(y)$, use $F_y = N$: $t^2 + h'(y) = t^2 + 4y$, so $h'(y) = 4y$, giving $h(y) = 2y^2$.

(a) Equilibria occur where $y' = 0$:

(b) Let $f(y) = 2y(2-3y)$. We test the sign in each region:

For $y < 0$: $f(-1) = 2(-1)(2+3) = -10 < 0$ (←)

For $0 < y < \dfrac{2}{3}$: $f(0.5) = 2(0.5)(2-1.5) = 0.5 > 0$ (→)

For $y > \dfrac{2}{3}$: $f(1) = 2(1)(2-3) = -2 < 0$ (←)

(c) At $y = 0$: arrows away → unstable. At $y = \dfrac{2}{3}$: arrows toward → stable.

(d) Starting at $y(0) = 0.5$, arrows point toward $\dfrac{2}{3}$.

Dividing by $(t+1)$ to get standard form:

Here $p(t) = \dfrac{-2}{t+1}$ and $g(t) = \dfrac{t^2}{t+1}$. Both have a discontinuity at $t = -1$.

Intervals of continuity: $(-\infty, -1)$ and $(-1, \infty)$. Since $t_0 = 0 \in (-1, \infty)$:

We have $M = 3t^{2}y + \cos t$ and $N = t^{3} - 2y$. Checking exactness:

Since $M_y = N_t$, the equation is exact. Integrating $M$ with respect to $t$:

Using $F_y = N$: $t^{3} + h'(y) = t^{3} - 2y$, so $h'(y) = -2y$, giving $h(y) = -y^2$.

General solution: $t^{3}y + \sin t - y^2 = C$. With $y(0) = 1$: $0 + 0 - 1 = -1$.

(a) Equilibria: $y = 0$ (double root), $y = 2$, $y = 3$.

Testing $f(y) = y^2(y-2)(y-3)$:

For $y < 0$: $f(-1) = (1)(-3)(-4) = 12 > 0$ (→)

For $0 < y < 2$: $f(1) = (1)(-1)(-2) = 2 > 0$ (→)

For $2 < y < 3$: $f(2.5) = (6.25)(0.5)(-0.5) < 0$ (←)

For $y > 3$: $f(4) = (16)(2)(1) = 32 > 0$ (→)

(b) $y=0$: semistable. $y=2$: stable. $y=3$: unstable.

(c) Starting at $y(0) = 2.5$ (between 2 and 3), arrows point toward 2.

(a) If $y_1 = 0$: $y_1' = 0$ and $3(0)^{2/3} = 0$. Both sides match. ✓

(b) Separating variables:

With $y(0) = 0$: $C = 0$. So $y^{1/3} = t$, giving $y_2(t) = t^3$.

(c) Computing the partial derivative:

At $y = 0$, this is undefined. Since $f_y$ is discontinuous there, the theorem doesn't apply—it makes no guarantee about uniqueness.

We have $M = ye^{ty} + 2t$ and $N = te^{ty} + 3y^{2}$. Using the product rule:

Since $M_y = N_t$, the equation is exact. Integrating $N$ with respect to $y$:

Using $F_t = M$: $ye^{ty} + g'(t) = ye^{ty} + 2t$, so $g'(t) = 2t$, giving $g(t) = t^2$.

(a) Logistic growth with harvesting:

(b) At steady state $P = 9$:

So $H = 0.405$ million per year.

(c) With $H = 0.405$, equilibria satisfy $0.18P(1 - P/12) = 0.405$:

Equilibria at $P = 3$ and $P = 9$. Testing $f(P) = 0.18P(1 - P/12) - 0.405$:

For $P < 3$: $f(0) = -0.405 < 0$ (←)

For $3 < P < 9$: $f(6) = 0.18(6)(0.5) - 0.405 = 0.135 > 0$ (→)

For $P > 9$: $f(12) = 0 - 0.405 < 0$ (←)

$P = 3$ is unstable, $P = 9$ is stable. With $P(0) = 6$, population increases to 9 million.

Note that $t^2 - 4 = (t-2)(t+2)$. In standard form:

Discontinuities: $t = -3$ (ln undefined), $t = -2, 1, 2$ (denominators). We need $t > -3$.

Intervals of continuity: $(-3, -2), (-2, 1), (1, 2), (2, \infty)$. Since $t_0 = 0 \in (-2, 1)$:

We have $M = \dfrac{2t}{t^2+y^2} + 1$ and $N = \dfrac{2y}{t^2+y^2} - 3y^{2}$. Using the quotient rule:

Since $M_y = N_t$, the equation is exact. Note that $\dfrac{\partial}{\partial t}[\ln(t^2+y^2)] = \dfrac{2t}{t^2+y^2}$. Integrating $M$ with respect to $t$:

Using $F_y = N$: $\dfrac{2y}{t^2+y^2} + h'(y) = \dfrac{2y}{t^2+y^2} - 3y^{2}$, so $h'(y) = -3y^{2}$, giving $h(y) = -y^3$.

With $y(1) = 0$: $\ln(1) + 1 - 0 = 1$.

(a) Setting $y' = 0$:

(b) Testing $f(y) = y(y-2)(y+2)$:

For $y < -2$: $f(-3) = (-3)(-5)(-1) = -15 < 0$ (←)

For $-2 < y < 0$: $f(-1) = (-1)(-3)(1) = 3 > 0$ (→)

For $0 < y < 2$: $f(1) = (1)(-1)(3) = -3 < 0$ (←)

For $y > 2$: $f(3) = (3)(1)(5) = 15 > 0$ (→)

$y = -2$: unstable. $y = 0$: stable. $y = 2$: unstable.

(c) With $y(0) = 1 \in (0, 2)$: arrows toward 0, so $\lim_{t\to\infty} y(t) = 0$.

(d) With $y(0) = -1 \in (-2, 0)$: arrows toward 0, so $\lim_{t\to\infty} y(t) = 0$.

Factoring: $t^2 - t - 6 = (t-3)(t+2)$. Discontinuities at $t = -2, 3$.

Intervals of continuity: $(-\infty, -2), (-2, 3), (3, \infty)$.

(a) $t_0 = 0 \in (-2, 3)$: $\boxed{(-2, 3)}$

(b) $t_0 = 5 \in (3, \infty)$: $\boxed{(3, \infty)}$

(c) $t_0 = -2$ is a discontinuity. The theorem does not apply at a point of discontinuity.

We have $M = e^{t}\sin y + 3y$ and $N = e^{t}\cos y + 3t - 2y$. Checking exactness:

Since $M_y = N_t$, the equation is exact. Integrating $M$ with respect to $t$:

Using $F_y = N$: $e^{t}\cos y + 3t + h'(y) = e^{t}\cos y + 3t - 2y$, so $h'(y) = -2y$, giving $h(y) = -y^2$.

With $y(0) = \dfrac{\pi}{2}$: $e^{0}\sin(\dfrac{\pi}{2}) + 0 - \dfrac{\pi^2}{4} = 1 - \dfrac{\pi^2}{4}$.

(a) Setting $3y(1-y/20) = 0$: equilibria at $y = 0$ and $y = 20$. Testing signs:

For $y < 0$: $f(-1) < 0$ (←)

For $0 < y < 20$: $f(10) > 0$ (→)

For $y > 20$: $f(25) < 0$ (←)

$y = 0$ is unstable, $y = 20$ is stable.

(b) Separating variables:

Partial fractions: $\dfrac{20}{y(20-y)} = \dfrac{1}{y} + \dfrac{1}{20-y}$. Integrating:

Exponentiating: $\dfrac{y}{20-y} = Ce^{3t}$. With $y(0) = 5$: $\dfrac{5}{15} = C$, so $C = \dfrac{1}{3}$.

Solving for $y$: $3y = (20-y)e^{3t}$, so $y(3 + e^{3t}) = 20e^{3t}$.

(c) As $t \to \infty$: $e^{-3t} \to 0$, so $y \to 20$. ✓

(a) Separating variables: $y\,dy = -4t\,dt$. Integrating:

(Choose the sign to match $y_0$.)

(b) For $y$ to be real: $y_0^2 - 4t^2 \geq 0$, i.e., $|t| \leq \dfrac{|y_0|}{2}$. Domain: $\left[-\dfrac{|y_0|}{2}, \dfrac{|y_0|}{2}\right]$.

(c) For the solution to exist on $[-1, 1]$: $\dfrac{|y_0|}{2} \geq 1$, so $|y_0| \geq 2$.

(a) $M = 2ty - 3$, $N = t^2 + 2$. We have $M_y = 2t$ and $N_t = 2t$. Exact.

Integrating $M$: $\int (2ty - 3)\,dt = t^2y - 3t + h(y)$.

From $F_y = N$: $t^2 + h'(y) = t^2 + 2$, so $h'(y) = 2$, giving $h(y) = 2y$.

(b) $M = y\cos(ty) + 2t$, $N = t\cos(ty) - 1$.

$M_y = \cos(ty) - ty\sin(ty)$ and $N_t = \cos(ty) - ty\sin(ty)$. Exact.

Integrating $N$ w.r.t. $y$: $\int (t\cos(ty) - 1)\,dy = \sin(ty) - y + g(t)$.

From $F_t = M$: $y\cos(ty) + g'(t) = y\cos(ty) + 2t$, so $g'(t) = 2t$, giving $g(t) = t^2$.

(c) $M = 3t^2 + y$, $N = t - 2y$. We have $M_y = 1$ and $N_t = 1$. Exact.

Integrating $M$: $\int (3t^2 + y)\,dt = t^3 + ty + h(y)$.

From $F_y = N$: $t + h'(y) = t - 2y$, so $h'(y) = -2y$, giving $h(y) = -y^2$.

Equilibria satisfy $y^2 - 4y + k = 0$, giving $y = 2 \pm \sqrt{4-k}$.

(a) $k=3$: $y = 2 \pm 1$, so equilibria at $y = 1$ and $y = 3$.

Testing $f(y) = (y-1)(y-3)$:

For $y < 1$: $f(0) = 3 > 0$ (→)

For $1 < y < 3$: $f(2) = -1 < 0$ (←)

For $y > 3$: $f(4) = 3 > 0$ (→)

$y=1$ is unstable, $y=3$ is stable.

(b) $k=4$: $y = 2$ (double root). $f(y) = (y-2)^2 \geq 0$. Arrows (→) everywhere except at $y = 2$.

$y = 2$ is semistable.

(c) $k=5$: $y = 2 \pm \sqrt{-1}$ gives complex roots. No real equilibria.

$f(y) = (y-2)^2 + 1 > 0$ for all real $y$. Since $y' > 0$ everywhere, all solutions increase without bound.

(a) $f(t,y) = t^2 + y^2$ and $f_y = 2y$. Both are continuous everywhere. At $(0,1)$: continuous. Yes.

(b) $f(t,y) = |y|$. The partial derivative $f_y = \text{sgn}(y)$ is undefined at $y = 0$. Since the initial condition has $y_0 = 0$, the theorem does not apply. No.

(c) $f(t,y) = \sqrt{|y|}$. The partial derivative $f_y \to \infty$ as $y \to 0$. Since the initial condition has $y_0 = 0$, the theorem does not apply. No.

(d) $f(t,y) = y\ln y$ (defined for $y > 0$) and $f_y = \ln y + 1$. At $(0, 1)$: $f = 0$ and $f_y = 1$, both defined and continuous. Yes.

(a) We have $M = 2ty + t^2y^2$ and $N = t^2 + \dfrac{2}{3}t^3y$. Checking exactness:

Since $M_y = N_t$, the equation is exact. ✓

(b) Integrating $N$ with respect to $y$:

Using $F_t = M$: $2ty + t^2y^2 + g'(t) = 2ty + t^2y^2$, so $g'(t) = 0$.

General solution: $t^2y + \dfrac{t^3y^2}{3} = C$.

(c) With $y(1) = 3$: $(1)(3) + \dfrac{1}{3}(1)(9) = 3 + 3 = 6$.