91影视

In problems 1-8 identify (do not solve) the equation as homogeneous, Bernoulli, linear coefficients, or of the form ${\mathbf{y}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{G}\left(\mathrm{ax}+\mathrm{by}\right)$.

$\left({\mathrm{ye}}^{-2x}+y^3\right)\mathbf{dx}\mathbf{-}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{x}}\mathbf{dy}\mathbf{=}\mathbf{0}$

In problems, 1-8 identify (do not solve) the equation as homogeneous, Bernoulli, linear coefficients, or of the form ${\mathbf{y}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{G}\left(\mathrm{ax}+\mathrm{by}\right)$.

$\mathbf{cos}\left(x+y\right)\mathbf{dy}\mathbf{=}\mathbf{sin}\left(x+y\right)\mathbf{dx}$

In problems 1-8 identify (do not solve) the equation as homogeneous, Bernoulli, linear coefficients, or of the form ${\mathbf{y}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{G}\left(\mathrm{ax}+\mathrm{by}\right)$.

$\left(y^3-{\mathrm{胃测}}^2\right)\mathbf{诲胃}\mathbf{+}\mathbf{2}{\mathbf{胃}}^{\mathbf{2}}\mathbf{ydy}\mathbf{=}\mathbf{0}$

Use the method discussed under 鈥淗omogeneous Equations鈥� to solve problems 9-16$鈥�\left(\mathbf{xy}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\right)鈥�\mathbf{dx}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}\mathbf{dy}\mathbf{=}\mathbf{0}$

Use the method discussed under 鈥淏ernoulli Equations鈥� to solve problems 21-28

$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathbf{y}\mathbf{=}{\mathbf{e}}^{\mathbf{x}}{\mathbf{y}}^{\mathbf{-}\mathbf{2}}$

In Problems 21鈥�26, solve the initial value problem.

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

Solve the initial value problem. \[\sqrt {\bf{y}} {\bf{dx + }}\left( {{\bf{1 + x}}} \right){\bf{dy = 0}}\,\,\,\,\,\,\,\,\,\,\,{\bf{y}}\left( {\bf{0}} \right){\bf{ = 1}}\]

Question: In Problems 1-30, solve the equation.

$\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{+}\mathbf{xy}\mathbf{=}\mathbf{0}$

For each of the following equations, find the most general function

so that the equation is exact.

$$\begin{gathered} \mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{M}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{dx}\mathbf{+}\left({\sec }^{2}y-\frac{x}{y}\right)\mathbf{dy}\mathbf{=}\mathbf{0} \\ \mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{M}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{dx}\mathbf{+}\mathbf{(}\mathbf{sin}\mathbf{}\mathbf{x}\mathbf{}\mathbf{cos}\mathbf{}\mathbf{y}\mathbf{-}\mathbf{xy}\mathbf{-}{\mathbf{e}}^{\mathbf{-}\mathbf{y}}\mathbf{)}\mathbf{dy}\mathbf{=}\mathbf{0} \end{gathered}$$

Question: Consider the initial value problem $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\sqrt{\mathbf{1}\mathbf{+}{\mathbf{sin}}^{\mathbf{2}}\mathbf{xy}}\mathbf{=}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}$.

(a)Using definite integration, show that the integrating factor for the differential equation can be written as $\mathbf{渭}\left(x\right)\mathbf{=}\left({鈭�}_0^x\sqrt{1+{\sin }^2\mathrm{tdt}}\right)$ and that the solution to the initial value problem is $\mathbf{y}\left(x\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{渭}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{\mathbf{鈭�}}_{\mathbf{0}}^{\mathbf{x}}\mathbf{渭}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{sds}\mathbf{+}\frac{\mathbf{2}}{\mathbf{渭}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}$

(b)Obtain an approximation to the solution at x= 1 by using numerical integration (such as Simpson鈥檚 rule, Appendix C) in a nested loop to estimate values of$\mathbf{渭}\left(x\right)$and, thereby, the value of${\mathbf{鈭�}}_{\mathbf{0}}^{\mathbf{1}}\mathbf{渭}\left(s\right)\mathbf{ds}$.

[Hint:First, use Simpson鈥檚 rule to approximate$\mathbf{渭}\left(x\right)$at x= 0.1, 0.2, . . . , 1. Then use these values and apply Simpson鈥檚 rule again to approximate${\mathbf{鈭�}}_{\mathbf{0}}^{\mathbf{1}}\mathbf{渭}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{ds}$]

(c)Use Euler鈥檚 method (Section 1.4) to approximate the solution at x= 1, with step sizes h= 0.1 and 0.05. [A direct comparison of the merits of the two numerical schemes in parts (b) and (c) is very complicated, since it should take into account the number of functional evaluations in each algorithm as well as the inherent accuracies.]