AP Calculus AB Exam Question:

Consider the following differential equation:

$\frac{dy}{dx} + y = y^2$

(a) By separating the variables, solve the given differential equation.

(b) Use the initial condition $y(0) = 1$ to find the particular solution.

Answer:

Part (a):

To solve the given differential equation $\frac{dy}{dx} + y = y^2$, we can separate the variables.

First, we write the equation as:

$\frac{dy}{dx} = y^2 - y$

Next, we can express the differential equation in terms of separate variables:

$\frac{dy}{y^2 - y} = dx$

To proceed with separation, we write the non-trivial terms as partial fractions. Factoring the denominator, we have:

$\frac{dy}{y(y-1)} = dx$

Now, we can rewrite the equation as:

$\frac{A}{y} + \frac{B}{y-1} = dx$

To find the values of A and B, we multiply both sides of the equation by the common denominator (y(y-1)):

$A(y-1) + By = dx$

Let's equate the coefficients of the constant term and y:

$A - A + By - B = dx$

$A + B = 0$ ...(1)

$-A = dx$ ...(2)

From equation (2), we can see that $A = -dx$.

Substituting this back into equation (1), we get:

$-dx + B = 0$

$B = dx$

Therefore, we have $A = -dx$ and $B = dx$.

Now, we rewrite the original equation, separating the variables:

$\int\left(\frac{-1}{y} + \frac{1}{y-1}\right)dy = \int dx$

Integrating both sides with respect to y, we have:

$-ln|y| + ln|y-1| = x + C$

Using the logarithmic properties, we can simplify the equation further:

$ln\left|\frac{y-1}{y}\right| = x + C$

Finally, we exponentiate both sides to eliminate the logarithms:

$\frac{y-1}{y} = e^{x+C}$

$\frac{y-1}{y} = Ce^x$

Cross-multiplying:

$y - 1 = Cye^x$

$y(1-Ce^x) = 1$

$y = \frac{1}{1-Ce^x}$

Therefore, the general solution to the given differential equation is:

$y = \frac{1}{1-Ce^x}$.

Part (b):

Now, we need to use the initial condition $y(0) = 1$ to find the particular solution.

Substituting $x = 0$ and $y = 1$ into the general solution, we have:

$1 = \frac{1}{1-Ce^0}$

$1 = \frac{1}{1-C}$

Solving for C, we have:

$1 - C = 1$

$C = 0$

Therefore, the particular solution to the initial condition is:

$y = \frac{1}{1-0e^x}$

$y = \frac{1}{1}$

$y = 1$

Hence, the particular solution to the differential equation with the initial condition $y(0) = 1$ is $y = 1$.

Therefore, the solution to the given differential equation with the initial condition is:

$y = \frac{1}{1-Ce^x}$

where $C = 0$.