Question:

Consider the differential equation:

(1) $ \frac{dy}{dx} = x^2y - 3x^2 $

(a) Use separation of variables to find the general solution of equation (1).

(b) Find the particular solution to equation (1) that satisfies the initial condition $y(1) = 4$.

Solution:

(a) General Solution:

To solve the given differential equation (1) using separation of variables, we start by isolating the variables. Rearranging equation (1), we have:

$ \frac{dy}{y-3} = x^2dx $

We will now integrate both sides of the equation with respect to their respective variables:

$ \int \frac{dy}{y-3} = \int x^2dx $

Integrating the left side with respect to y gives:

$ \ln|y-3| = \frac{x^3}{3} + C_1 $

where $ C_1 $ is the constant of integration.

To solve for y, we exponentiate both sides:

$ |y-3| = e^{\frac{x^3}{3} + C_1} $

Using the absolute value, we can separate it into two cases:

(Case 1) $ y-3 > 0 $:

$ y - 3 = e^{\frac{x^3}{3} + C_1} $

Simplifying gives:

$ y = 3 + e^{\frac{x^3}{3} + C_1} $

where $ C_1 $ is an arbitrary constant.

(Case 2) $ y-3 < 0 $:

$ -(y - 3) = e^{\frac{x^3}{3} + C_1} $

Simplifying gives:

$ y = 3 - e^{\frac{x^3}{3} + C_1} $

where $ C_1 $ is an arbitrary constant.

Thus, the general solution to differential equation (1) is:

$ y = \begin{cases}3 + e^{\frac{x^3}{3} + C_1} & \text{if } y > 3\ 3 - e^{\frac{x^3}{3} + C_1} & \text{if } y < 3\end{cases} $

where $ C_1 $ is an arbitrary constant.

(b) Particular Solution with Initial Condition:

To find the particular solution to equation (1) that satisfies the initial condition $ y(1) = 4 $, we substitute the initial condition into the general solution obtained in part (a).

Since $ y(1) = 4 > 3 $, we consider the case $ y > 3 $. Plugging in $ x = 1 $ and $ y = 4 $ into the general solution, we have:

$ 4 = 3 + e^{\frac{1^3}{3} + C_1} $

Simplifying this equation, we find:

$ e^{\frac{1}{3} + C_1} = 1 $

Taking the natural logarithm of both sides, we get:

$ \frac{1}{3} + C_1 = \ln(1) = 0 $

Therefore:

$ C_1 = -\frac{1}{3} $

Plugging $ C_1 $ back in, we obtain:

$ y = 3 + e^{\frac{x^3}{3} - \frac{1}{3}} $

Hence, the particular solution to equation (1) that satisfies the initial condition $ y(1) = 4 $ is:

$ y = 3 + e^{\frac{x^3}{3} - \frac{1}{3}} $

Please note that in case $ y < 3 $, the initial condition $ y(1) = 4 $ would not hold as it implies $ y < 3 $. Therefore, we do not consider that case.

Answer: The particular solution is $ y = 3 + e^{\frac{x^3}{3} - \frac{1}{3}} $.