RaySix

Post

at December 5th 2023, 8:16:38 pm.

Sure, here is an AP Calculus AB exam question on the topic of separation of variables:

Question:

Consider the first-order differential equation given by

\frac{dy}{dx} = \frac{4x^2 + 1}{2y}

Part (a):

By using separation of variables, solve the given differential equation.

Part (b):

Find the particular solution that passes through the point $(1, 2)$ .

Answer:

Part (a):

To solve the differential equation using separation of variables, we start by multiplying both sides by $2y$ to get:

2y \cdot \frac{dy}{dx} = 4x^2 + 1

Next, we can separate the variables to solve the equation as follows:

2y \cdot dy = (4x^2 + 1) \cdot dx

Integrating both sides:

\int 2y \cdot dy = \int (4x^2 + 1) \cdot dx

y^2 = \frac{4x^3}{3} + x + C

Where $C$ is the constant of integration.

Part (b):

To find the particular solution, we use the initial condition $(1, 2)$ to solve for the constant $C$ :

2^2 = \frac{4\cdot 1^3}{3} + 1 + C

4 = \frac{4}{3} + 1 + C

C = 4 - \frac{4}{3} - 1 = \frac{8}{3} - 1 = \frac{5}{3}

So, the particular solution passing through $(1, 2)$ is given by:

y^2 = \frac{4x^3}{3} + x + \frac{5}{3}

Therefore, the particular solution is:

y = \sqrt{\frac{4x^3}{3} + x + \frac{5}{3}}

This completes the solution.