$\dfrac{dy}{dx} = 7x$

$\dfrac{dy}{dx}$ $= \dfrac{2x}{4 + x^2}$

$\dfrac{dy}{dx} \cdot \dfrac{1}{x}$ $= 7x^2 + 5$

$\dfrac{dy}{dx} = \dfrac{3y}{7x}$

$\dfrac{dy}{dx} = − \dfrac{x}{y}$

$\dfrac{dy}{dx} = xy^2$

$\dfrac{dy}{dx} = x^2y$

$\dfrac{dy}{dx} = \dfrac{y}{1 + x^2},$ $\text{ and }$ $y(0) = 1$

$\dfrac{dy}{dx} = 7y\tan(x)$, $y(0) = 128$

$\dfrac{dy}{dx} = e^{2x}$, $y(0) = 0$

$y” − y’ − 30y = 0$

$\text{Identify}$ $\text{if}$ $y = e^{−5x}$ $\text{is}$ $\text{a}$ $\text{solution}$ $\text{to}$ $\text{the}$ $\text{differential}$ $\text{equation.}$

$y”’ + 12y” + 47y’ + 60y = 0$

$\text{Identify}$ $\text{if}$ $y = e^{−6x}$ $\text{is}$ $\text{a}$ $\text{solution}$ $\text{to}$ $\text{the}$ $\text{differential}$ $\text{equation.}$

$y” − 4y’ + 4y = 0$

$\text{Identify}$ $\text{if}$ $y = xe^{2x}$ $\text{is}$ $\text{a}$ $\text{solution}$ $\text{to}$ $\text{the}$ $\text{differential}$ $\text{equation.}$

$x^2y” + xy’ = 0$

$\text{Confirm}$ $\text{that}$ $y = \ln(x)$ $\text{is}$ $\text{a}$ $\text{solution}$ $\text{to}$ $\text{the}$ $\text{differential}$ $\text{equation.}$

$y^{(4)} − y = 0$

$\text{Identify}$ $\text{if}$ $y = \cos(x)$ $\text{is}$ $\text{a}$ $\text{solution}$ $\text{to}$ $\text{the}$ $\text{differential}$ $\text{equation.}$