AP Calculus AB Exam Question:

Consider the differential equation:

$\frac{{dy}}{{dx}} = 2xy$

(a) Use separation of variables to find the general solution of the differential equation.

(b) Find the particular solution of the differential equation that satisfies the initial condition $y(0) = 1$.

Answer:

(a) To solve the differential equation using separation of variables, we first rewrite it as:

$\frac{{dy}}{{dx}} = 2xy$

Dividing both sides by $y$ and multiplying both sides by $dx$, we have:

$\frac{{dy}}{{y}} = 2x\,dx$

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

$\int \frac{{dy}}{{y}} = \int 2x\,dx$

Integrating, we get:

$\ln|y| = x^2 + C_1$

Where $C_1$ is the constant of integration.

Taking exponential of both sides to eliminate the logarithm, we have:

$y = e^{x^2 + C_1}$

Simplifying, we obtain:

$y = Ce^{x^2}$

Where $C = e^{C_1}$ is an arbitrary constant.

Thus, the general solution of the given differential equation is $y = Ce^{x^2}$.

(b) To find the particular solution that satisfies the initial condition $y(0) = 1$, we substitute the given values into the general solution:

$1 = Ce^{0^2} \Rightarrow 1 = C\cdot e^0 \Rightarrow 1 = C$

Hence, the particular solution that satisfies the initial condition is $y = e^{x^2}$.

Note: The constant of integration, $C$, is found by substituting the initial condition into the general solution.