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\(\tan\) is discontinuous at odd multiples of \(\frac{\pi}{2}\), since \(\frac{\pi}{2} < \pi < \frac{3\pi}{2}\), the interval is \((\frac{\pi}{2}, \frac{3\pi}{2})\).

Dividing both sides by \(\ln(t)\) yields \begin{equation*} y' + \frac{y}{\ln(t)} = \frac{\cot(t)}{\ln(t)} \end{equation*} for \(\ln(t) \neq 0 \iff t \neq 1\). \(\cot(t)\) forces out \(t\) to be in the range \((0, \pi)\). By finding the intersection of those constraints, we get an interval \((1, \pi)\).

Based on the direction field and on the differential equation, for \(y_0 < 0\), the slopes eventually become negative, therefore tend to \(-\infty\). If \(y_0=0\), then we get an equilibrium solution. Note that slopes are zero along the curves \(y=0\) and \(ty = 3\).

Solutions with \(t_{0}<0\) all tend to \(-\infty\). Solutions with initial conditions \(\left(t_{0}, y_{0}\right)\) to the right of the parabola \(t=1+y^{2}\) asymptotically approach the parabola as \(t \rightarrow \infty\). Integral curves with initial conditions above the parabola (and \(\left.y_{0}>0\right)\) also approach the curve. The slopes for solutions with initial conditions below the parabola (and \(\left.y_{0}<0\right)\) are all negative. These solutions tend to \(-\infty\).

The solution of the initial value problem \begin{equation*} y_1'+2y_1=0, \quad y_1(0) = 1 \end{equation*} is \(y_1(t) = e^{-2t}\). Therefore by approaching to \(1\) from the left side (\(1^-\) notation), we get \(y(1^-) = y_1(1) = e^{-2}\). On the interval \((1, \infty)\), the differential equation is \(y_2'+y_2=0\) with \(y_2(t)=ce^{-t}\). Therefore by approaching \(1\) from the right side (notationally \(1^+\)), we see \(y(1^+)=y_2(1)=ce^{-1}\). Equating both the limits of the function \begin{align*} y(1^-) = y(1^+) \iff c = e^{-1} \end{align*} Therefore the global solution is \begin{equation*} y(t) = \begin{cases} e^{-2t}, \quad 0 \leq t \leq 1\\ e^{-1-t}, \quad t > 1 \end{cases} \end{equation*}

The Eleventh Edition (latest) of the book doesn’t have this problem.

They have the form \(M(x,y) + N(x,y) \frac{dy}{dx} = 0\). So \begin{align*} M(x,y) = 3x^2-2xy+2 \quad \text{and} \quad N(x,y) = 6y^2-x^2+3 \end{align*} Then we see \(\frac{\partial M}{\partial y} = -2x\) and \(\frac{\partial N}{\partial x} = -2x\). Therefore, our equation is of exact form. So our solution \(F_x = M \implies F = \int M dx = x^3 - x^2y + 2x + g(y)\). Then \begin{equation*} F_y = -x^2+g'(y) = N \implies g'(y) = 6y^2+3 \implies g(y)=2y^3 + 3y \end{equation*} Finally, \begin{equation*} F = x^3 - x^2y + 2x +2y^3 + 3y = C \end{equation*}

\begin{align*} \frac{dy}{dx} = - \frac{ax-by}{bx-cy} \\ \iff (ax-by)dx + (bx-cy)dy = 0 \end{align*} Now, \(M = ax-by\) and \(N = bx -cy\). See that \begin{align*} M_y = -b \neq N_x = b \end{align*} The differential equation is not exact.

Integrating \(\psi_{y}=N\), while holding \(x\) constant, yields \(\psi(x, y)=\int N(x, y) d y+h(x)\) Taking the partial derivative with respect to \(x, \psi_{x}=\int \frac{\partial}{\partial x} N(x, y) d y+h^{\prime}(x)\) . Now set \(\psi_{x}=M(x, y)\) and therefore \(h^{\prime}(x)=M(x, y)-\int \frac{\partial}{\partial x} N(x,y) dy\). Based on the fact that \(M_{y}=N_{x}\), it follows that \(\frac{\partial}{\partial y}\left[h^{\prime}(x)\right]=0\). Hence the expression for \(h^{\prime}(x)\) can be integrated to obtain \begin{align*} h(x)=\int M(x, y) d x-\int\left[\int \frac{\partial}{\partial x} N(x, y) d y\right] d x \end{align*}

\begin{align*} M = x^2y^3,\quad \quad N = x(1+y^2)\\ \implies M_y = 3x^2y^2, \quad \quad N_x = 1+y^2 \end{align*} Trivially, not exact. Let \(\mu(x,y) = \frac{1}{xy^3}\), then \begin{align*} M\times\mu = x, \quad \quad N\times\mu = \frac{1+y^2}{y^3} \implies (M\times\mu)_y = 0, \quad \quad (N\times\mu)_x = 0 \end{align*} Now they’re exact!

So then just find that \(F = \frac{x^2}{2} - \frac{1}{2y^2}+\ln(y)\)

\begin{align*} M = 3x^2y+2xy+y^3,\quad \quad N = x^2+y^2\\ \implies M_y = 3x^2+2x+3y^2, \quad \quad N_x = 2x \end{align*} Let us find the integrating factor \begin{align*} \mu(y) &= \exp\left(\int \frac{M_y-N_x}{N} dx\right)\\ &= \exp\left(\int \frac{3x^2+2x+3y^2-2x}{x^2+y^2} dx\right)\\ &= \exp\left(\int 3 dx\right)\\ &= e^{3x} \end{align*} Simply confirm that \(M\mu\) and \(N\mu\) are now exact. Find \(F(x,y) = e^{3x}y(3x^2+y^2) = C\)