We are given the following initial value problem to solve:

$\frac{dy}{dx} + xy^2 e^{-x} = -xe^{-x}, \quad y > 0, \quad y(0) = 3.$

Step-by-Step Solution

We can solve this using an integrating factor or by attempting to simplify the equation directly.

1. First, simplify the differential equation:

   Move all terms involving $y$ and $\frac{dy}{dx}$ to one side: $\frac{dy}{dx} = -xe^{-x} - xy^2 e^{-x}.$ Factor out the common term $e^{-x}$: $\frac{dy}{dx} = -xe^{-x}(1 + y^2).$

2. Separate the variables:

   To solve, separate the $y$-terms and $x$-terms: $\frac{dy}{1 + y^2} = -xe^{-x} dx.$

3. Integrate both sides:

   The left-hand side can be integrated as an inverse tangent function: $\int \frac{dy}{1 + y^2} = \tan^{-1}(y).$

   The right-hand side requires integration by parts. Let’s solve: $\int -xe^{-x} dx.$

   Using integration by parts, we set $u = x$ and $dv = e^{-x} dx$. Then $du = dx$ and $v = -e^{-x}$. Applying the integration by parts formula: $\int -xe^{-x} dx = -xe^{-x} + \int e^{-x} dx = -xe^{-x} - e^{-x}.$

   So, the integral of the right-hand side is: $\int -xe^{-x} dx = -e^{-x}(x + 1).$

4. Write the integrated equation:

   Now we have the integrated equation: $\tan^{-1}(y) = -e^{-x}(x + 1) + C,$ where $C$ is the constant of integration.

5. Apply the initial condition:

   Use the initial condition $y(0) = 3$ to solve for $C$. Substituting $x = 0$ and $y = 3$ into the equation: $\tan^{-1}(3) = -e^{0}(0 + 1) + C,$ $\tan^{-1}(3) = -1 + C,$ $C = \tan^{-1}(3) + 1.$

6. Final solution:

   The final solution is: $\tan^{-1}(y) = -e^{-x}(x + 1) + \tan^{-1}(3) + 1.$

To express $y$ explicitly, take the tangent of both sides: $y = \tan\left(-e^{-x}(x + 1) + \tan^{-1}(3) + 1\right).$

Would you like further details on any of the steps or clarifications?

Related Questions:

1. What is the general method for solving first-order separable differential equations?
2. Can you explain the integration by parts technique in more detail?
3. How do you find integrating factors for linear first-order differential equations?
4. What are the conditions for the existence and uniqueness of solutions for initial value problems?
5. How can inverse trigonometric functions help in solving differential equations?

Tip:

Always check if a differential equation can be simplified or transformed using known techniques like separation of variables or integrating factors.