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Decide whether the statement made is True or False. The function $\mathbf{y}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{)}$ is a solution to $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\frac{\mathbf{y}\mathbf{-}\mathbf{1}}{\mathbf{x}\mathbf{+}\mathbf{3}}$.

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

If it is, then solve it.

$\mathbf{(}\mathbf{2}\mathbf{xy}\mathbf{+}\mathbf{3}\mathbf{)}\mathbf{dx}\mathbf{+}\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{dy}\mathbf{=}\mathbf{0}$

Let $蠒\mathbf{\left(}\mathbf{x}\mathbf{\right)}$denote the solution to the initial value problem

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

猞� Show that $蠒\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{-}蠒\mathbf{\text{'}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{-}\mathbf{x}\mathbf{+}蠒\mathbf{\left(}\mathbf{x}\mathbf{\right)}$

猞� Argue that the graph of $蠒$ is decreasing for x near zero and that as x increases from zero, $蠒\mathbf{\left(}\mathbf{x}\mathbf{\right)}$decreases until it crosses the line y = x, where its derivative is zero.

猞� Let x* be the abscissa of the point where the solution curve $\mathbf{y}\mathbf{=}蠒\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ crosses the line $y=x$.Consider the sign of $蠒\mathbf{(}\mathbf{x}\mathbf{*}\mathbf{)}$ and argue that $蠒$ has a relative minimum at x*.

猞� What can you say about the graph of $\mathbf{y}\mathbf{=}蠒\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ for x > x*?

猞� Verify that y = x 鈥� 1 is a solution to $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{y}$ and explain why the graph of $\mathbf{y}\mathbf{=}蠒\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ always stays above the line $\mathbf{y}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{1}$.

猞� Sketch the direction field for $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{y}$ by using the method of isoclines or a computer software package.

猞� Sketch the solution $\mathbf{y}\mathbf{=}蠒\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ using the direction field in part (f).

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.

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

Decide whether the statement made is True or False. The relation ${\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{3}}\mathbf{-}{\mathbf{e}}^{\mathbf{y}}\mathbf{=}\mathbf{1}$ is an implicit solution to $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\frac{{\mathbf{e}}^{\mathbf{y}}\mathbf{-}\mathbf{2}\mathbf{x}}{\mathbf{3}{\mathbf{y}}^{\mathbf{2}}}$.

Find a general solution for the given differential equation.

(a)${\mathit{y}}^{\mathbf{\left(}\mathbf{4}\mathbf{\right)}}\mathbf{+}\mathbf{2}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{4}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{2}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{3}\mathit{y}\mathbf{=}\mathbf{0}$

(b)${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{3}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{5}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathit{y}\mathbf{=}\mathbf{0}$

(c)${\mathit{y}}^{\mathbf{\left(}\mathbf{5}\mathbf{\right)}}\mathbf{-}{\mathit{y}}^{\mathbf{\left(}\mathbf{4}\mathbf{\right)}}\mathbf{+}\mathbf{2}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{2}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{-}\mathit{y}\mathbf{=}\mathbf{0}$

(d)${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{2}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{=}{\mathit{e}}^{\mathbf{x}}\mathbf{+}\mathit{x}$