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### Messages - David Chan

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1
##### Term Test 1 / Re: P-4
« on: February 13, 2018, 11:15:40 PM »
The characteristic equation for $\mathcal{L}[y] = y'' + 2y' + 2y$ is $r^2 + 2r + 2 = 0$ which has roots $r = -1 \pm i$.  Thus, the general solution to the corresponding homogeneous equation $\mathcal{L}[y_c] = 0$ is $$y_c = c_1e^{-t}\cos(t) + c_2e^{-t}\sin(t)$$
We now look for a particular solution to the non-homogeneous equation $\mathcal{L}[y] = -e^{-t} + 10\cos(t)$ by method of undetermined coefficients.
First, we look for $Y_1 = Ae^{-t}$ such that $\mathcal{L}[Y_1] = -e^{-t}$.  Differentiating, we get $Y_1' = -Ae^{-t}$ and $Y_1'' = Ae^{-t}$.  Plugging these back into the differential equation, we get $$Ae^{-t} - 2Ae^{-t} + 2Ae^{-t} = Ae^{-t} = -e^{-t} \quad \implies \quad A = -1$$
Next, we look for $Y_2 = B\cos(t) + C\sin(t)$ such that $\mathcal{L}[Y_2] = 10\cos(t)$.  Differentiating, we get $Y_2' = -B\sin(t) + C\cos(t)$ and $Y_2'' = -B\cos(t) - C\sin(t)$.  Plugging these back into the differential equation, we get $$-B\cos(t) - C\sin(t) - 2B\sin(t) + 2C\cos(t) + 2B\cos(t) + 2C\sin(t) = (B + 2C)\cos(t) + (-2B + C)\sin(t)$$ $$\implies \begin{cases} \ \ B + 2C &= 10 \\ -2B + C &= 0 \end{cases} \quad \implies \quad B = 2,\ C = 4$$
So, we have $Y_1 = -e^{-t}$ and $Y_2 = 2\cos(t) + 4\sin(t)$.  Thus, the general solution to the differential equation is $$\boxed{y = c_1e^{-t}\cos(t) + c_2e^{-t}\sin(t) -e^{-t} + 2\cos(t) + 4\sin(t)}$$

2
##### Term Test 1 / Re: P-3
« on: February 13, 2018, 10:50:31 PM »
(a) The characteristic equation for $\mathcal{L}[y] = y'' + y' - 2y$ is $r^2 + r - 2 = 0$ which has roots $r_1 = 1$ and $r_2 = -2$.  Thus, the general solution to the corresponding homogeneous equation $\mathcal{L}[y_c] = 0$ is $$y_c = c_1e^t + c_2e^{-2t}$$
We now look for a particular solution to the non-homogeneous equation $\mathcal{L}[Y] = -6 + 9e^{-2t}$ by method of undetermined coefficients.
First, we look for $Y_1 = A$ such that $\mathcal{L}[Y_1] = -6$.  Differentiating, we get $Y_1' = Y_1'' = 0$.  Plugging these back into the differential equation, we get $$-2A = -6 \implies A = 3$$
Note that $e^{-2t}$ is already part of the homogeneous solution, so instead, we look for $Y_2 = Bte^{-2t}$ such that $\mathcal{L}[Y_2] = 9e^{-2t}$.  Differentiating, we get $Y_2' = Be^{-2t} -2Bte^{-2t}$ and $Y_2'' = 4Bte^{-2t} - 4Be^{-2t}$.  Plugging these back into the differential equation, we get $$4Bte^{-2t} - 4Be^{-2t} + Be^{-2t} -2Bte^{-2t} - 2Bte^{-2t} = -3Be^{-2t} = 9e^{-2t} \quad \implies\quad B = -3$$
So, we have $Y_1 = 3$ and $Y_2 = -3te^{-2t}$.  Thus, the general solution to the differential equation is $$\boxed{y = c_1e^t + c_2e^{-2t} - 3te^{-2t} + 3}$$

(b) Note that $y'(t) = c_1 e^t - 2 c_2 e^{-2 t} + 6 t e^{-2 t} - 3 e^{-2 t}$.  Given the initial conditions $y(0) = 0$ and $y'(0) = 0$, we find that $$0 = c_1 + c_2 + 3$$ $$0 = c_1 - 2c_2 - 3$$
Solving, this system of equations, we get $c_1 = -1$ and $c_2 = -2$, so our solution is $$\boxed{y = -e^t - 2e^{-2t} - 3te^{-2t} + 3}$$

3
##### Term Test 1 / Re: P-1
« on: February 13, 2018, 10:35:50 PM »
Let $M(x, y) = y^2$ and $N(x, y) = 3xy - \cos(y)$.  Then, $$\frac{\partial}{\partial y}M(x, y) = 2y \qquad \text{ and } \qquad \frac{\partial}{\partial x}N(x, y) = 3y$$
We can see that this equation is not exact, however, note that $\frac{N_x - M_y}{M} = \frac{3y - 2y}{y^2} = \frac{1}{y}$ is a function of $y$ only.  Thus, there is an integrating factor $\mu(y)$ that satisfies the differential equation $$\frac{\mathrm{d}\mu}{\mathrm{d}y} = \left(\frac{N_x - M_y}{M}\right)\mu = \frac{1}{y}\mu \qquad \implies \qquad \mu(y) = y$$
Multiplying our original equation by $\mu(y)$, we have $$(y^3) + (3xy^2 - y\cos(y))\frac{\mathrm{d}y}{\mathrm{d}x} = 0$$
We can see that this equation is exact, since $\frac{\partial}{\partial y}(y^3) = 3y^2 = \frac{\partial}{\partial x} 3xy^2 - y\cos(y)$.  Thus, there exists a function $\psi(x, y)$ such that \begin{align*}\psi_x(x, y) &= y^3 \tag{1} \\\psi_y(x, y) &= 3xy^2 - y\cos(y) \tag{2}\end{align*}
Integrating (1) with respect to $x$, we get \begin{align*}\psi(x, y) &= \int y^3 \,\mathrm{d}x = xy^3 + h(y)\end{align*}
for some function $h$ of $y$.  Next, differentiating with respect to $y$, and equating with (2), we get $$\psi_y(x, y) = 3xy^2 + h'(y) = 3xy^2 - y\cos(y)$$
Therefore, $$h'(y) = -y\cos(y) \quad \implies \quad h(y) = -y\sin(y) - \cos(y) \quad \implies \quad \psi(x, y) = xy^3 - y\sin(y) - \cos(y)$$
Thus, the solutions of the differential equation are given implicitly by $$\boxed{xy^3 - y\sin(y) - \cos(y) = C}$$
Given the initial value $y(1) = \pi$, we have $\pi^3 + 1 = C$, so the particular solution is $$\boxed{xy^3 - y\sin(y) - \cos(y) = \pi^3 + 1}$$

4
##### Quiz-2 / Re: Q2-T0901
« on: February 02, 2018, 02:34:11 PM »
We want to find an integrating factor $\mu$ as a function of $xy$ such that $(\mu M)_y = (\mu N)_x$.  Let $z = xy$.  Thus, $\mu(xy) = \mu(z(x, y))$  Then $$\mu_x(xy) = \frac{\mathrm{d}\mu}{\mathrm{d}z} \frac{\partial z}{\partial x} = y\frac{\mathrm{d}\mu}{\mathrm{d}z} \qquad \text{ and } \qquad \mu_y(xy) = \frac{\mathrm{d}\mu}{\mathrm{d}z} \frac{\partial z}{\partial y} = x\frac{\mathrm{d}\mu}{\mathrm{d}z}$$ Therefore, \begin{align*}
(\mu M)_y = (\mu N)_x & \implies \mu M_y + xM\frac{\mathrm{d}\mu}{\mathrm{d}z} = \mu N_x + yN\frac{\mathrm{d}\mu}{\mathrm{d}z}\\
&\implies \mu(M_y - N_x) = \frac{\mathrm{d}\mu}{\mathrm{d}z}(yN - xM)\\
&\implies \frac{\mathrm{d}\mu}{\mathrm{d}z} = \mu\left(\frac{N_x - M_y}{xM - yN}\right)
\end{align*}
Therefore, $$\mu(z) = \exp\left(\int R(z)\,\mathrm{d}z\right) \qquad \text{ where } R(z) = R(xy) = \frac{N_x - M_y}{xM - yN}$$
Returning to our original differential equation, let $$M(x, y) = 3x + \frac{6}{y} \qquad \text{ and } \qquad N(x, y) = \frac{x^2}{y} + 3 \frac{y}{x} = 0$$  Then  $$\frac{\partial}{\partial y}M(x, y) = \frac{-6}{y^2} \qquad \text{ and } \qquad \frac{\partial}{\partial x}N(x, y) = \frac{2x}{y} - \frac{3y}{x^2}$$  We can see that this equation is not exact, however, note that $$\frac{N_x - M_y}{xM - yN} = \frac{\frac{2x}{y} - \frac{3y}{x^2} + \frac{6}{y^2}}{x\left(3x + \frac{6}{y}\right) - y\left(\frac{x^2}{y} + 3 \frac{y}{x}\right)} = \frac{\frac{2x}{y} - \frac{3y}{x^2} + \frac{6}{y^2}}{2x^2 + \frac{6x}{y} - \frac{3y^2}{x}} = \frac{\frac{2x}{y} - \frac{3y}{x^2} + \frac{6}{y^2}}{xy\left(\frac{2x}{y} - \frac{3y}{x^2} + \frac{6}{y^2}\right)} = \frac{1}{xy}$$
Thus, we have an integrating factor $$\mu(xy) = \exp\left(\int \frac{1}{z}\,\mathrm{d}z\right) = e^{\log{|z|}} = z = xy$$
Multiplying the original differential equation through by our integrating factor, we have $$\left(3x^2y + 6x\right) + \left(x^3 + 3y^2\right)\frac{\mathrm{d}y}{\mathrm{d}x} = 0$$ We can see that this differential equation is exact because $$\frac{\partial}{\partial y}(3x^2y + 6x) = 3x^2 = \frac{\partial}{\partial x}(x^3 + 3y^2)$$  Thus, there exists a function $\psi(x, y)$ such that \begin{align*}\psi_x(x, y) &= 3x^2y + 6x \tag{1} \\\psi_y(x, y) &= x^3 + 3y^2 \tag{2}\end{align*}
Integrating (1) with respect to $x$, we get $$\psi(x, y) = x^3y + 3x^2 + h(y)$$ for some function $h$ of $y$. Next, differentiating with respect to $y$, and equating with (2), we get $$\psi_y(x, y) = x^3 + h'(y)$$
Therefore, $$h'(y) = 3y^2 \qquad \implies \qquad h(y) = y^3$$ and we have $$\psi(x, y) = x^3y + 3x^2 + y^3$$
Thus, the solutions of the differential equation are given implicitly by $$x^3y + 3x^2 + y^3 = C$$

5
##### Quiz-2 / Re: Q2-T0701
« on: February 02, 2018, 02:33:25 PM »
Let $$M(x, y) = 3x^2y + 2xy + y^3 \qquad \text{ and } \qquad N(x, y) = x^2 + y^2$$
Then, $$\frac{\partial}{\partial y}M(x, y) = 3x^2 + 2x + 3y^2 \qquad \text{ and } \qquad \frac{\partial}{\partial x}N(x, y) = 2x$$
We can see that this equation is not exact, however, note that $$\frac{M_y - N_x}{N} = \frac{3(x^2 + y^2)}{x^2 + y^2} = 3$$ is a function of $x$ only.  Thus, there is an integrating factor $\mu(x)$ that satisfies the differential equation $$\frac{\mathrm{d}\mu}{\mathrm{d}x} = \left(\frac{M_y - N_x}{N}\right)\mu = 3\mu \qquad \implies \qquad \mu = e^{3x}$$
Multiplying our original equation by $\mu(x)$, we have $$(3x^2e^{3x}y + 2xe^{3x}y + e^{3x}y^3) + e^{3x}(x^2 + y^2)\frac{\mathrm{d}y}{\mathrm{d}x} = 0$$
We can see that this equation is exact, since $$\frac{\partial}{\partial y}(3x^2e^{3x}y + 2xe^{3x}y + e^{3x}y^3) = 3x^2e^{3x} + 2xe^{3x} + 3e^{3x}y^2 = \frac{\partial}{\partial x} e^{3x}(x^2 + y^2)$$
Thus, there exists a function $\psi(x, y)$ such that \begin{align*}\psi_x(x, y) &= 3x^2e^{3x}y + 2xe^{3x}y + e^{3x}y^3 \tag{1} \\\psi_y(x, y) &= e^{3x}(x^2 + y^2) \tag{2}\end{align*}
Integrating (1) with respect to $x$, we get \begin{align*}\psi(x, y) &= \int e^{3x}(3x^2y + 2xy + y^3)\,\mathrm{d}x \\&= e^{3x}\left(x^2y + \frac{1}{3}y^3\right) + h(y) \tag*{by parts}\end{align*}
for some function $h$ of $y$.  Next, differentiating with respect to $y$, and equating with (2), we get $$\psi_y(x, y) = e^{3x}(x^2 + y^2) + h'(y) = e^{3x}(x^2 + y^2)$$
Therefore, $$h'(y) = 0 \implies h \text{ is constant }$$  Taking $h(y) = 0$, we get $$\psi(x, y) = e^{3x}\left(x^2y + \frac{1}{3}y^3\right)$$
Thus, the solutions of the differential equation are given implicitly by $$e^{3x}\left(x^2y + \frac{1}{3}y^3\right) = C$$

6
##### Quiz-2 / Re: Q2-T0601
« on: February 02, 2018, 02:30:56 PM »
Let $$M(x, y) = 9x^2 + y - 1 \qquad \text{ and } \qquad N(x, y) = -4y + x$$
Then, $$\frac{\partial}{\partial y}M(x, y) = 1 \qquad \text{ and } \qquad \frac{\partial}{\partial x}N(x, y) = 1$$
(Also note that $M$, $N$, $M_y$, $N_x$ are all continuous)  Since $M_y = N_x$, the equation is exact.  Therefore, there exists a function $\psi(x, y)$ such that \begin{align*}\psi_x(x, y) &= 9x^2 + y - 1 = M \tag{1} \\\psi_y(x, y) &= -4y + x = N \tag{2}\end{align*}
Integrating (1) with respect to $x$, we get $$\psi(x, y) = 3x^3 + xy - x + h(y)$$ for some function $h$ of $y$.  Next, differentiating with respect to $y$, and equating with (2), we get $$\psi_y(x, y) = x + h'(y) = -4y + x$$
Therefore, $$h'(y) = -4y \implies h(y) -2y^2$$
and we have $$\psi(x, y) = 3x^3 + xy - x - 2y^2$$
By our choice of $\psi$, we know that $$\frac{\partial \psi}{\partial x} = M \qquad \text{ and } \qquad \frac{\partial \psi}{\partial y} = N$$ so we can rewrite our original differential equation as $$\frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial y} \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}}{\mathrm{d}x}(\psi(x, y)) = 0$$  Therefore, the (implicit) general solution to out differential equation is $$\psi(x, y) = -2y^2 + xy + 3x^3 - x = C$$
We can solve this explicitly for $y$ to using the quadratic formula to get $$y = \frac{x \pm \sqrt{x^2 + 8(3x^3 - x - C)}}{4}$$
Given our initial condition, $y(1) = 0$, we have $$3(1)^3 - 1 = C \implies C = 2$$
Thus, we have our solution $$y = \frac{x - \sqrt{24x^3 + x^2 - 8x - 16}}{4}, \qquad \text{ valid for } \ 24x^3 + x^2 - 8x - 16 > 0 \iff x > .985$$

7
##### Quiz-2 / Re: Q2-T0501
« on: February 02, 2018, 02:29:51 PM »
Let $$M(x, y) = (x + 2)\sin(y) \qquad \text{ and } \qquad N(x, y) = x\cos(y)$$
Then, $$\frac{\partial}{\partial y}M(x, y) = (x + 2)\cos(y) \qquad \text{ and } \qquad \frac{\partial}{\partial x}N(x, y) = \cos(y)$$
Note that $M_y \neq N_x$, so the equation is not exact.  However, multiplying through by $\mu(x, y) = xe^x$, we get a new equation $$(x + 2)xe^x\sin(y) + x^2e^x\cos(y)y' = 0$$
We can see that this equation is exact, since $$\frac{\partial}{\partial y} (x + 2)xe^x\sin(y)= (x + 2)xe^x\cos(y) = \frac{\partial}{\partial x} x^2e^x\cos(y)$$
Therefore, there exists a function $\psi(x, y)$ such that \begin{align*}\psi_x(x, y) &= (x + 2)xe^x\sin(y) \tag{1} \\\psi_y(x, y) &= x^2e^x\cos(y) \tag{2}\end{align*}
Integrating (1) with respect to $x$, we get $$\psi(x, y) = \int (x + 2)xe^x\sin(y)\,\mathrm{d}x = x^2e^x\sin(y) + h(y)$$
for some function $h$ of $y$.  Next, differentiating with respect to $y$, and equating with (2), we get $$\psi_y(x, y) = x^2e^x\cos(y) + h'(y) = x^2e^x\cos(y)$$
Therefore, $$h'(y) = 0 \implies h \text{ is constant }$$  Taking $h(y) = 0$, we get $$\psi(x, y) = x^2e^x\sin(y)$$
Thus, the solutions of the differential equation are given implicitly by $$x^2e^x\sin(y) = C$$

8
##### Quiz-2 / Re: Q2-T0401
« on: February 02, 2018, 02:28:58 PM »
Let $$M(x, y) = 1 \qquad \text{ and } \qquad N(x, y) = \left(\frac{x}{y} - \sin(y)\right)$$
Then, $$\frac{\partial}{\partial y}M(x, y) = 0 \qquad \text{ and } \qquad \frac{\partial}{\partial x}N(x, y) = \frac{1}{y}$$
We can see that this equation is not exact, however, note that $$\frac{N_x - M_y}{M} = \frac{1}{y}$$ is a function of $y$ only.  Thus, there is an integrating factor $\mu(y)$ that satisfies the differential equation $$\frac{\mathrm{d}\mu}{\mathrm{d}y} = \left(\frac{N_x - M_y}{M}\right)\mu = \frac{\mu}{y} \qquad \implies \qquad \mu = y$$
Multiplying our original equation by $\mu(y)$, we have $$y + (x - y\sin(y))y' = 0$$
We can see that this equation is exact, since $$\frac{\partial}{\partial y}(y) = 1 = \frac{\partial}{\partial x}(x - y\sin(y))$$
Thus, there exists a function $\psi(x, y)$ such that \begin{align*}\psi_x(x, y) &= y \tag{1} \\\psi_y(x, y) &= x - y\sin(y) \tag{2}\end{align*}
Integrating (1) with respect to $x$, we get $$\psi(x, y) = xy + h(y)$$ for some arbitrary function $h$ of $y$.  Next, differentiating with respect to $y$, and equating with (2), we get $$\psi_y(x, y) = x + h'(y)$$
Therefore, $$h'(y) = -y\sin(y) \qquad \implies \qquad h(y) = -\int y \sin(y)\,\mathrm{d}y = y\cos(y) - \sin(y)$$
and we have $$\psi(x, y) = xy + y\cos(y) - \sin(y)$$
Thus, the solutions of the differential equation are given implicitly by $$xy + y\cos(y) - \sin(y) = C$$

9
##### Quiz-2 / Re: Q2-T0301
« on: February 02, 2018, 02:28:17 PM »
Let $$M(x, y) = 2x - y \qquad \text{ and } \qquad N(x, y) = 2y - x$$
Then, $$\frac{\partial}{\partial y}M(x, y) = -1 \qquad \text{ and } \qquad \frac{\partial}{\partial x}N(x, y) = -1$$
(Also note that $M$, $N$, $M_y$, $N_x$ are all continuous)  Since $M_y = N_x$, the equation is exact.  Therefore, there exists a function $\psi(x, y)$ such that \begin{align*}\psi_x(x, y) &= 2x - y = M \tag{1} \\\psi_y(x, y) &= 2y - x = N \tag{2}\end{align*}
Integrating (1) with respect to $x$, we get $$\psi(x, y) = x^2 - xy + h(y)$$ for some function $h$ of $y$.  Next, differentiating with respect to $y$, and equating with (2), we get $$\psi_y(x, y) = -x + h'(y) = 2y - x$$
Therefore, $$h'(y) = 2y \implies h(y) = y^2$$
and we have $$\psi(x, y) = x^2 - xy + y^2$$
By our choice of $\psi$, we know that $$\frac{\partial \psi}{\partial x} = M \qquad \text{ and } \qquad \frac{\partial \psi}{\partial y} = N$$ so we can rewrite our original differential equation as $$\frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial y} \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}}{\mathrm{d}x}(\psi(x, y)) = 0$$  Therefore, the (implicit) general solution to out differential equation is $$\psi(x, y) = x^2 - xy + y^2 = C$$
We can solve this explicitly for $y$ to using the quadratic formula to get $$y = \frac{x \pm \sqrt{-3x^2 + 4C}}{2}$$
Given the initial condition $y(1) = 3$, we have $$(1)^2 + (1)(3) - (3)^2 = C \implies C = 7$$
Thus, we have our solution $$y = \frac{x + \sqrt{28 - 3x^2}}{2}, \qquad \text{ valid for \vert x \vert < \frac{\sqrt{28}}{3}}$$

10
##### Quiz-2 / Re: Q2-T0201
« on: February 02, 2018, 02:27:28 PM »
Let $$M(x, y) = 3x^2y + 2xy + y^3 \qquad \text{ and } \qquad N(x, y) = x^2 + y^2$$
Then, $$\frac{\partial}{\partial y}M(x, y) = 3x^2 + 2x + 3y^2 \qquad \text{ and } \qquad \frac{\partial}{\partial x}N(x, y) = 2x$$
We can see that this equation is not exact, however, note that $$\frac{M_y - N_x}{N} = \frac{3(x^2 + y^2)}{x^2 + y^2} = 3$$ is a function of $x$ only.  Thus, there is an integrating factor $\mu(x)$ that satisfies the differential equation $$\frac{\mathrm{d}\mu}{\mathrm{d}x} = \left(\frac{M_y - N_x}{N}\right)\mu = 3\mu \qquad \implies \qquad \mu = e^{3x}$$
Multiplying our original equation by $\mu(x)$, we have $$(3x^2e^{3x}y + 2xe^{3x}y + e^{3x}y^3) + e^{3x}(x^2 + y^2)\frac{\mathrm{d}y}{\mathrm{d}x} = 0$$
We can see that this equation is exact, since $$\frac{\partial}{\partial y}(3x^2e^{3x}y + 2xe^{3x}y + e^{3x}y^3) = 3x^2e^{3x} + 2xe^{3x} + 3e^{3x}y^2 = \frac{\partial}{\partial x} e^{3x}(x^2 + y^2)$$
Thus, there exists a function $\psi(x, y)$ such that \begin{align*}\psi_x(x, y) &= 3x^2e^{3x}y + 2xe^{3x}y + e^{3x}y^3 \tag{1} \\\psi_y(x, y) &= e^{3x}(x^2 + y^2) \tag{2}\end{align*}
Integrating (1) with respect to $x$, we get \begin{align*}\psi(x, y) &= \int e^{3x}(3x^2y + 2xy + y^3) \\&= e^{3x}\left(x^2y + \frac{1}{3}y^3\right) + h(y) \tag*{by parts}\end{align*}
for some function $h$ of $y$.  Next, differentiating with respect to $y$, and equating with (2), we get $$\psi_y(x, y) = e^{3x}(x^2 + y^2) + h'(y) = e^{3x}(x^2 + y^2)$$
Therefore, $$h'(y) = 0 \implies h \text{ is constant }$$  Taking $h(y) = 0$, we get $$\psi(x, y) = e^{3x}\left(x^2y + \frac{1}{3}y^3\right)$$
Thus, the solutions of the differential equation are given implicitly by $$e^{3x}\left(x^2y + \frac{1}{3}y^3\right) = C$$

11
##### Quiz-2 / Re: Q2-T0101
« on: February 02, 2018, 02:26:19 PM »
Let $$M(x, y) = x^2 y^3 \qquad \text{ and } \qquad N(x, y) = x(1 + y^2)$$
Then, $$\frac{\partial}{\partial y}M(x, y) = 3x^2y^2 \qquad \text{ and } \qquad \frac{\partial}{\partial x}N(x, y) = 1 + y^2$$
Note that $M_y \neq N_x$, so the equation is not exact.  Note also that $y \equiv 0$ is a solution.  Supposing $x, y \neq 0$, we can multiply through by $\mu(x, y) = \frac{1}{xy^3}$, we get a new equation $$x + \frac{1 + y^2}{y^3}y' = 0$$
We can see that this equation is exact, since $$\frac{\partial}{\partial y}(x) = 0 = \frac{\partial}{\partial x}\left(\frac{1 + y^2}{y^3}\right)$$
Therefore, there exists a function $\psi(x, y)$ such that \begin{align*}\psi_x(x, y) &= x \tag{1} \\\psi_y(x, y) &= \frac{1 + y^2}{y^2} \tag{2}\end{align*}
Integrating (1) with respect to $x$, we get $$\psi(x, y) = \frac{1}{2}x^2 + h(y)$$ for some function $h$ of $y$.  Next, differentiating with respect to $y$, and equating with (2), we get $$\psi_y(x, y) = h'(y) = \frac{1 + y^2}{y^3}$$
Therefore, $$h'(y) = \frac{1 + y^2}{y^3} \implies h(y) = \int \frac{\mathrm{d}y}{y^3} + \int \frac{\mathrm{d}y}{y}= \log{|y|} - \frac{1}{2y^2}$$
and we have $$\psi(x, y) = \frac{1}{2}x^2 + \log{|y|} - \frac{1}{2y^2}$$
Thus, the solutions of the differential equation are given implicitly by $$\frac{1}{2}x^2 + \log{|y|} - \frac{1}{2y^2} = C \qquad \text{ or } \qquad y \equiv 0$$

12
##### Quiz-1 / Re: Q1-T0101
« on: January 25, 2018, 09:50:30 AM »
This is a first order, linear differential equation.  So, it can be solved by method of integrating factor.  First, $$\mu(t) = \exp\left(\int 3\,dt\right) = e^{3t}$$  So, multiplying by the integrating factor, we get that $$\frac{d}{dt}(e^{3t}y) = te^{3t} + e^{t}$$  Integrating both sides, \begin{align*}ye^{3t} &= \int 3e^{3t} dt + \int e^t dt\\ &= \frac{t}{3}e^{3t} - \frac{1}{9}e^{3t} + e^{t} + C\end{align*} Therefore, $$y = \frac{t}{3} - \frac{1}{9} + e^{-2t} + Ce^{-3t}$$ is the general solution.  As $$t \to \infty, \qquad y \to \frac{t}{3} - \frac{1}{9}$$

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##### Web Bonus Problems / Re: Web bonus problem--Week 3
« on: January 22, 2018, 02:59:02 PM »
My (typed) solution:

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