AP - Calculus AB/BC:: Differential Equations

A differential equation is best described as:

- **A)** An equation relating two constants
- **B)** An equation involving an unknown function and one or more of its derivatives
- **C)** An equation that can only be solved numerically, never exactly
- **D)** An equation with no derivatives, only algebraic terms

To verify that \(y = e^{2x}\) is a solution to the differential equation \(y' = 2y\), what should you do?

- **A)** Verification is impossible without solving the equation from scratch
- **B)** Integrate \(y\) and check if it equals \(2y\)
- **C)** Simply substitute \(x=0\) and check that \(y=1\)
- **D)** Differentiate \(y=e^{2x}\) to get \(y'=2e^{2x}\), then check that this equals \(2y = 2e^{2x}\) -- confirming they match

Verify whether \(y = 3e^{-x}\) is a solution to \(y' + y = 0\).

- **A)** Yes, since \(y' = -3e^{-x}\), and \(y'+y = -3e^{-x}+3e^{-x} = 0\)
- **B)** Cannot be determined without an initial condition
- **C)** No, since \(y'+y\) is never zero for exponential functions
- **D)** No, since \(y'=3e^{-x}\)

In separation of variables, the general strategy for solving \(\dfrac{dy}{dx} = g(x)h(y)\) is:

- **A)** Differentiate both sides again
- **B)** Immediately apply the Chain Rule without rearranging
- **C)** This technique only works for linear differential equations
- **D)** Algebraically rearrange to get all \(y\)-terms (with \(dy\)) on one side and all \(x\)-terms (with \(dx\)) on the other, then integrate both sides separately

Use separation of variables to find the general solution to \(\dfrac{dy}{dx} = \dfrac{x}{y}\).

- **A)** \(y = x + C\)
- **B)** \(xy = C\)
- **C)** \(y^2 = x^2 + C\)
- **D)** \(y = \dfrac{x^2}{2} + C\)

Use separation of variables to find the general solution to \(\dfrac{dy}{dx} = ky\) (the fundamental exponential growth/decay equation).

- **A)** \(y = k e^{Cx}\)
- **B)** \(y = C + kx\)
- **C)** \(y = Ckx\)
- **D)** \(y = Ce^{kx}\)

A population \(P(t)\) grows according to \(\dfrac{dP}{dt} = 0.03P\), with \(P(0) = 500\). Find the particular solution \(P(t)\).

- **A)** \(P(t) = 500e^{0.03t}\)
- **B)** \(P(t) = 0.03e^{500t}\)
- **C)** \(P(t) = 500(0.03)^t\)
- **D)** \(P(t) = 500 + 0.03t\)

How is an 'initial condition' used when solving a differential equation?

- **A)** It determines the differential equation's degree
- **B)** It is a given value (like \(y(x_0)=y_0\)) used to solve for the arbitrary constant \(C\) in the general solution, producing one specific 'particular solution'
- **C)** It is only relevant for first-order equations, never used in practice
- **D)** It replaces the need to integrate at all