91影视

Using the method of isoclines sketch the direction field for\[{\bf{y = - }}\frac{{{\bf{4x}}}}{{\bf{y}}}\].

Show that $\mathbf{蠒}\left(x\right)\mathbf{=}{\mathbf{Ce}}^{\mathbf{3}\mathbf{x}}\mathbf{+}\mathbf{1}$is a solution tolocalid="1663944867164" $\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{=}\mathbf{-}\mathbf{3}$for any choice of the constant C. Thus,${\mathbf{Ce}}^{\mathbf{3}\mathbf{x}}\mathbf{+}\mathbf{1}$ is a one-parameter family of solutions to the differential equation. Graph several of the solution curves using the same coordinate axes.

In the electrical circuit of Figure\({\bf{5}}{\bf{.52}}\), take\({{\bf{R}}_{\bf{1}}}{\bf{ = }}{{\bf{R}}_{\bf{2}}}{\bf{ = 1\Omega ,C = 1\;F}}\), and\({\bf{L = 1H}}\). Derive three equations for the unknown currents\({{\bf{I}}_{\bf{1}}}{\bf{,}}{{\bf{I}}_{\bf{2}}}\), and \({{\bf{I}}_{\bf{3}}}\) by writing Kirchhoff's voltage law for loops \({\bf{1}}\)and\({\bf{2}}\), and Kirchhoff's current law for the top juncture. Find the general solution.

The direction field for the equation\[\frac{{{\bf{dp}}}}{{{\bf{dt}}}}{\bf{ = }}\frac{{{\bf{p(2p - 1)(p - 3)}}}}{{\bf{9}}}\],where pis the population (in thousands) at time tof a certain species, is plotted in Figure 1.17.

(a)If the initial population is 2500, what can you say about the limiting population as \[{\bf{t}} \to \infty \]?

(b)If the initial population is 4000, will it ever decrease to 3500?

(c)If p(1)= 0.3, what is \[\mathop {{\bf{lim}}\;}\limits_{{\bf{t}} \to \infty } {\bf{p(t)}}\]?

Let c >0. Show that the function $\mathbf{蠒}\left(x\right)\mathbf{=}{\left(c^2-x^2\right)}^{\mathbf{-}\mathbf{1}}$is a solution to the initial value problem$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{2}{\mathbf{xy}}^{\mathbf{2}}\mathbf{,}\mathbf{y}\left(0\right)\mathbf{=}\frac{\mathbf{1}}{{\mathbf{c}}^{\mathbf{2}}}\mathbf{,}$on the interval$\mathbf{-}\mathbf{c}\mathbf{<}\mathbf{x}\mathbf{<}\mathbf{c}$. Note that this solution becomes unbounded as x approaches $\mathbf{卤}\mathit{c}$. Thus, the solution exists on the interval$\left(-未,未\right)$ with $\mathit{未}\mathbf{=}\mathit{c}$, but not for larger$\mathit{未}$. This illustrates that in Theorem 1, the existence interval can be quite small (IFC is small) or quite large (if c is large). Notice also that there is no clue from the equation$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{2}{\mathbf{xy}}^{\mathbf{2}}$ itself, or from the initial value, that the solution will 鈥渂low up鈥� at$\mathit{x}\mathbf{=}\mathbf{卤}\mathit{c}$.

In Problem 19, solve the given initial value problem

$\begin{array}{l}y\text{'}\text{'}\text{'}鈭�y\text{'}\text{'}鈭�4y\text{'}+4y=0\\ y\left(0\right)=鈭�4\\ y\text{'}\left(0\right)=鈭�1\\ y\text{'}\text{'}\left(0\right)=鈭�19\end{array}$

In Problems 9鈥�20, determine whether the equation is exact.

If it is, then solve it.

$\left(\mathbf{2}\mathbf{x}\mathbf{+}\frac{\mathbf{y}}{\mathbf{1}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}}\right)\mathbf{dx}\mathbf{+}\left(\frac{\mathbf{x}}{\mathbf{1}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}}\mathbf{-}\mathbf{2}\mathbf{y}\right)\mathbf{dy}\mathbf{=}\mathbf{0}$

Show that the equation${\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathbf{=}\mathbf{0}$ has no (real-valued) solution.

In problems 1-4 Use Euler鈥檚 method to approximate the solution to the given initial value problem at the points $\mathbf{x}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{,}\mathbf{0}\mathbf{.}\mathbf{3}\mathbf{,}\mathbf{0}\mathbf{.}\mathbf{4}$, and $\mathbf{0}\mathbf{.}\mathbf{5}$, using steps of size $\mathbf{0}\mathbf{.}\mathbf{1}\left(\mathbf{h}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}\right)$.

$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\frac{\mathbf{-}\mathbf{x}}{\mathbf{y}}$,$\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{4}$

In Problems , identify the equation as separable, linear, exact, or having an integrating factor that is a function of either x alone or y alone.

$\left(\mathbf{2}\mathbf{x}\mathbf{+}{\mathbf{yx}}^{\mathbf{-}\mathbf{1}}\right)\mathbf{dx}\mathbf{+}\left(\mathbf{xy}\mathbf{-}\mathbf{1}\right)\mathbf{dy}\mathbf{=}\mathbf{0}$