Introduction

In Problems 3–8, determine whether the given function is a solution to the given differential equation.

$\mathbf{x}\mathbf{=}\mathbf{2}\mathbf{ }\mathbf{cos}\mathbf{ }\mathbf{t}\mathbf{-}\mathbf{3}\mathbf{ }\mathbf{sin}\mathbf{ }\mathbf{t}\mathbf{ }\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{x}\mathbf{=}\mathbf{0}$

In Problems 3–8, determine whether the given function is a solution to the given differential equation.

$\mathbf{\theta }\mathbf{=}\mathbf{2}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}\mathbf{-}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}$, $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{\theta }}{{\mathbf{dt}}^{\mathbf{2}}}\mathbf{-}\mathbf{\theta }\frac{\mathbf{d\theta }}{\mathbf{dt}}\mathbf{+}\mathbf{3}\mathbf{\theta }\mathbf{=}\mathbf{-}\mathbf{2}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}$

In Problems 3-8, determine whether the given function is a solution to the given differential equation.

$\mathbf{x}\mathbf{=}\mathbf{cos}\mathbf{ }\mathbf{2}\mathbf{t}$, $\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{+}\mathbf{tx}\mathbf{=}\mathbf{sin}\mathbf{ }\mathbf{2}\mathbf{t}$

In Problems 3–8, determine whether the given function is a solution to the given differential equation.

$\mathbf{y}\mathbf{=}{\mathbf{e}}^{\mathbf{2}\mathbf{x}}\mathbf{-}\mathbf{3}{\mathbf{e}}^{\mathbf{-}\mathbf{x}}$, $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{-}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{0}$

Question: In Problems 3–8, determine whether the given function is a solution to the given differential equation.

$\mathbf{y}\mathbf{=}\mathbf{3}\mathbf{ }\mathbf{sin}\mathbf{ }\mathbf{2}\mathbf{x}\mathbf{+}{\mathbf{e}}^{\mathbf{-}\mathbf{x}}$, $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{5}{\mathbf{e}}^{\mathbf{-}\mathbf{x}}$

In Problems 9-13, determine whether the given relation is an implicit solution to the given differential equation. Assume that the relationship implicitly defines y as a function of x and use implicit differentiation.

${\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{4}$, $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\frac{\mathbf{x}}{\mathbf{y}}$

In Problems 9-13, determine whether the given relation is an implicit solution to the given differential equation. Assume that the relationship implicitly defines y as a function of x and use implicit differentiation.

$\mathbf{y}\mathbf{-}{\mathbf{log}}_{\mathbf{e}}\mathbf{y}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{1}$, $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\frac{\mathbf{2}\mathbf{xy}}{\mathbf{y}\mathbf{-}\mathbf{1}}$

In Problems 9–13, determine whether the given relation is an implicit solution to the given differential equation. Assume that the relationship implicitly defines y as a function of x and use implicit differentiation.

${\mathbf{e}}^{\mathbf{xy}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{1}$, $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\frac{{\mathbf{e}}^{\mathbf{-}\mathbf{xy}}\mathbf{-}\mathbf{y}}{{\mathbf{e}}^{\mathbf{-}\mathbf{xy}}\mathbf{+}\mathbf{x}}$

In Problems 9-13, determine whether the given relation is an implicit solution to the given differential equation. Assume that the relationship implicitly defines y as a function of x and use implicit differentiation.

$\mathbf{sin} \mathbf{y}\mathbf{+}\mathbf{xy}\mathbf{-}{\mathbf{x}}^{\mathbf{3}}\mathbf{=}\mathbf{2}$, $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\frac{\mathbf{6}\mathbf{xy}\mathbf{\text{'}}\mathbf{+}{\left(\mathbf{y}\mathbf{\text{'}}\right)}^{\mathbf{3}}\mathbf{sin} \mathbf{y}\mathbf{-}\mathbf{2}{\left(\mathbf{y}\mathbf{\text{'}}\right)}^{\mathbf{2}}}{\mathbf{3}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{y}}$

Show that $\varphi \left(\mathbf{x}\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}} \mathbf{sin} \mathbf{x}\mathbf{+}{\mathbf{c}}_{\mathbf{2}} \mathbf{cos} \mathbf{x}\mathbf{,}$ is a solution to $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0}$ for any choice of the constants ${\mathbf{c}}_{\mathbf{1}}$ and ${\mathbf{c}}_2$. Thus, ${\mathbf{c}}_{\mathbf{1}} \mathbf{sin} \mathbf{x}\mathbf{+}{\mathbf{c}}_{\mathbf{2}} \mathbf{cos} \mathbf{x}\mathbf{,}$ is a two-parameter family of solutions to the differential equation.

Let c >0. Show that the function $\mathbf{\varphi }\left(x\right)\mathbf{=}{\left(c^2-x^2\right)}^{\mathbf{-}\mathbf{1}}$ 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 $\mathbf{\pm }\mathit{c}$. Thus, the solution exists on the interval $\left(-\delta ,\delta \right)$ with $\mathit{\delta }\mathbf{=}\mathit{c}$, but not for larger $\mathit{\delta }$. 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 “blow up” at $\mathit{x}\mathbf{=}\mathbf{\pm }\mathit{c}$.

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.

Determine which values of m the function $\mathbf{\varphi }\left(x\right)\mathbf{=}{\mathbf{e}}^{\mathbf{mx}}$ is a solution to the given equation.

(a) $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{+}\mathbf{6}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathbf{5}\mathbf{y}\mathbf{=}\mathbf{0}$

(b) $\frac{{\mathbf{d}}^{\mathbf{3}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{3}}}\mathbf{+}\mathbf{3}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{+}\mathbf{2}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{0}$

In Problems 23-28, determine whether Theorem 1 implies that the given initial value problem has a unique solution.

$\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{-}\mathbf{ty}\mathbf{=}{\mathbf{sin}}^{\mathbf{2}}\mathbf{t}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{y}\left(\pi \right)\mathbf{=}\mathbf{5}$

In Problems 23-28, determine whether Theorem 1 implies that the given initial value problem has a unique solution.

$\mathbf{3}\mathbf{x}\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{+}\mathbf{4}\mathbf{t}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{ }\mathbf{x}\left(2\right)\mathbf{=}\mathbf{-}\mathbf{\pi }$

In Problems 23-28, determine whether Theorem 1 implies that the given initial value problem has a unique solution.

$\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{+}\mathbf{cos}\mathbf{ }\mathbf{x}\mathbf{=}\mathbf{sin}\mathbf{ }\mathbf{t}\mathbf{,}\mathbf{ }\mathbf{x}\left(\pi \right)\mathbf{=}\mathbf{0}$

In Problems 23-28, determine whether Theorem 1 implies that the given initial value problem has a unique solution.

$\mathbf{y}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{x}\mathbf{,}\mathbf{ }\mathbf{y}\left(1\right)\mathbf{=}\mathbf{0}$

In Problems 23-28, determine whether Theorem 1 implies that the given initial value problem has a unique solution.

$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{3}\mathbf{x}\mathbf{-}\sqrt[\mathbf{3}]{\mathbf{y}\mathbf{-}\mathbf{1}}\mathbf{,}\mathbf{ }\mathbf{y}\left(2\right)\mathbf{=}\mathbf{1}$

Consider the question of Example 5 $\mathbf{y}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{=}\mathbf{0}$

1. Does Theorem 1 imply the existence of a unique solution to (13) that satisfies $\mathbf{y}\left(x_0\right)\mathbf{=}\mathbf{0}$?
2. Show that when ${\mathbf{x}}_{\mathbf{0}}\mathbf{\ne }\mathbf{0}$ equation (13) can’t possibly have a solution in a neighbourhood of $\mathbf{x}\mathbf{=}{\mathbf{x}}_{\mathbf{0}}$ that satisfies $\mathbf{y}\left(x_0\right)\mathbf{=}\mathbf{0}$.
3. Show two distinct solutions to (13) satisfying $\mathbf{y}\left(0\right)\mathbf{=}\mathbf{0}$ ( See Figure 1.4 on page 9).

The initial value problem \[\frac{{{\bf{dx}}}}{{{\bf{dt}}}}{\bf{ = 3}}{{\bf{x}}^{\frac{{\bf{2}}}{{\bf{3}}}}}{\bf{,}}\;{\bf{x(0) = 1}}\] has a unique solution in some open interval around t = 0.

The initial value problem \[\frac{{{\bf{dx}}}}{{{\bf{dt}}}}{\bf{ = 3}}{{\bf{y}}^{\frac{{\bf{2}}}{{\bf{3}}}}}{\bf{,}}\;{\bf{y(2) = 1}}{{\bf{0}}^{{\bf{ - 100}}}}\]has a unique solution in some open interval around x = 2.

The solution to the initial value problem \[\frac{{{\bf{dy}}}}{{{\bf{dx}}}}{\bf{ = (x - 2)(y - 3}}{{\bf{)}}^{\bf{2}}}{\bf{,}}\;{\bf{y(0) = 0}}\], will always be less than 3; that is, \[{\bf{y(x) < 3}}\] for\[{\bf{x}} \ge 0\].

Use Euler’s method with step size \[{\bf{h = }}\frac{{\bf{1}}}{{\bf{2}}}\] to approximate the solution to the initial value problem \[\frac{{{\bf{dy}}}}{{{\bf{dx}}}}{\bf{ = x - }}{{\bf{y}}^{\bf{2}}}{\bf{,}}\;{\bf{y(1) = 2}}\] at x = 2.

If Euler’s method with step size\[{\bf{h = }}\frac{{\bf{1}}}{{\bf{n}}}\], where n is a positiveinteger, is used to approximate the solution to the initial value problem\[\frac{{{\bf{dy}}}}{{{\bf{dx}}}}{\bf{ + y = 0,y(0) = 1}}\] what formula (expressed as a function of n) do you obtain for the approximation of y (1 ) ? What is the exact value of y (1 )?

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

The direction field for the equation\[\frac{{{\bf{dp}}}}{{{\bf{dt}}}}{\bf{ = }}\frac{{{\bf{p(2p - 1)(p - 3)}}}}{{\bf{9}}}\],where p is the population (in thousands) at time t of 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 \[\mathop {{\bf{lim}}\;}\limits_{{\bf{t}} \to \infty } {\bf{p(t)}}\]?