Post

Created by @nathanedwards
 at October 31st 2023, 9:57:45 pm.

AP Calculus AB Exam Question - Separation of Variables:

Find the general solution to the following differential equation using separation of variables:

dydx=3x−1\frac{dy}{dx} = \frac{3}{x-1}

Solution:

To solve this differential equation, we will use the method of separation of variables. The first step is to write the equation in the form dydx=g(x)⋅f(y)\frac{dy}{dx} = g(x) \cdot f(y), where g(x)g(x) represents the function of xx and f(y)f(y) represents the function of yy.

For our given equation, we can rewrite it as:

dydx=3x−1\frac{dy}{dx} = \frac{3}{x-1}

Let's multiply both sides by dxdx to separate the variables:

dy=3x−1⋅dxdy = \frac{3}{x-1} \cdot dx

Now, we separate the variables by moving dxdx to one side and dydy to the other side:

dy=3x−1⋅dxdy = \frac{3}{x-1} \cdot dx

Next, we integrate both sides of the equation with respect to their respective variables:

∫dy=∫3x−1 dx\int dy = \int \frac{3}{x-1} \, dx

The integral of dydy is simply yy, and we can evaluate the integral of 1x−1\frac{1}{x-1} using the natural logarithm function:

y=3∫1x−1 dxy = 3 \int \frac{1}{x-1} \, dx

To evaluate this integral, we use the substitution method. Let u=x−1u = x-1, then du=dxdu = dx. Substituting uu and dudu into the equation, we have:

y=3∫1u duy = 3 \int \frac{1}{u} \, du

Integrating 1u\frac{1}{u} with respect to uu gives us ln⁡∣u∣+C\ln|u| + C, where CC is the constant of integration:

y=3ln⁡∣u∣+C=3ln⁡∣x−1∣+Cy = 3 \ln|u| + C = 3 \ln|x-1| + C

Finally, we have the general solution to the differential equation:

y=3ln⁡∣x−1∣+Cy = 3 \ln|x-1| + C

where CC is the constant of integration.