Question

Solve the initial value problem $$ y^{\prime}=2 y^{2}+x y^{2}, \quad y(0)=1 $$ and determine where the solution attains its minimum value.

Short answer

Based on the solution above, the minimum value of the solution is attained when x = -2, and at this point, y(-2) = 1/5.

Step-by-step solution

Separate the variables y and x

First, we need to rewrite the given equation to separate the variables y and x. The given equation is: $$ y^{\prime}=2y^2 + xy^2 $$ Divide both sides by \(y^2\): $$ \frac{y^{\prime}}{y^2} = 2 + x $$

Integrate both sides

Now that the equation is separable, we can integrate both sides with respect to x: $$ \int\frac{y^{\prime}}{y^2} dx = \int (2 + x) dx $$ This equation can be written as: $$ \int\frac{dy}{y^2} = \int (2 + x) dx $$ Now, integrate both sides: $$ -\frac{1}{y} = 2x + \frac{1}{2}x^2 + C $$

Apply the initial condition

Apply the initial condition y(0) = 1: $$ -\frac{1}{1} = 2(0) + \frac{1}{2}(0)^2 + C $$ So, the constant C = -1, and the particular solution is: $$ -\frac{1}{y} = 2x + \frac{1}{2}x^2 -1 $$ Rewrite this equation for y: $$ y = \frac{1}{1 - 2x - \frac{1}{2}x^2} $$

Determine the minimum value

To find the minimum value of the solution, we have to find the critical points of the function. It's important to note that the critical points occur when the derivative is either equal to zero or undefined. Differentiate y with respect to x: $$ \frac{dy}{dx} = -\frac{1}{(1 - 2x - \frac{1}{2}x^2)^2}(-2 - x) $$ Set the derivative equal to zero: $$ 0 = -\frac{1}{(1 - 2x - \frac{1}{2}x^2)^2}(-2 - x) $$ Since the numerator is always nonzero, the derivative can only equal zero when the denominator is zero: $$ 0 = -2 - x $$ The critical point occurs at x = -2. Thus, the solution attains its minimum value when x = -2. To find the minimum value, plug the critical point into the function for y: $$ y(-2) = \frac{1}{1 - 2(-2) - \frac{1}{2}(-2)^2} $$ The minimum value attained by the solution is: $$ y(-2) = \frac{1}{5}

Key concepts

Separable Differential Equations

Separable differential equations are a class of differential equations that can be separated into two parts, one involving only the independent variable (often denoted as x) and the other involving only the dependent variable (often denoted as y). This separation allows us to solve for y by using simple integration techniques.

For example, the initial value problem
\[ y' = 2y^2 + xy^2, \quad y(0) = 1 \]
involves a separable differential equation. By dividing both sides of the equation by y^2, the equation becomes:
\[ \frac{y'}{y^2} = 2 + x \]
This separation of variables allows us to integrate both sides, ultimately leading us to the solution of the differential equation. The process simplifies complex differential equations into more manageable forms. Integrating both sides of the equation with respect to x after separation gives us the relationship that defines y implicitly or explicitly in terms of x.

Integrating Factors

Integrating factors are a technique used to solve linear non-separable differential equations. An integrating factor is a function that is multiplied by the differential equation to make it possible to write the equation as an exact differential, which can then be integrated easily.

However, in our example, the equation is already separable and hence we do not require an integrating factor in this case. Nevertheless, the concept of integrating factors is of vital importance, especially when the given differential equation cannot be separated as easily. It's worth noting that the integrating factor is generally a function of x or y and is derived from the equation itself. This tool turns an intractable equation into a form where the methods of calculus can be applied straightforwardly to find a solution.

Critical Points

Critical points of a function are locations on its graph where the derivative is zero or the derivative does not exist. They are crucial in determining the function's relative maximums, minimums, and points of inflection.

Once we find the solution to the differential equation, as we did for
\[ y = \frac{1}{1 - 2x - \frac{1}{2}x^2} \]
we then differentiate it to find the critical points. In our specific case, the critical points occur when the derivative of the function with respect to x is zero or undefined. We calculated the derivative and set it equal to zero, then found the critical point at x = -2. This critical point corresponds to the minimum value of the original function, and by evaluating y at x = -2, we determined that the minimum value is \( \frac{1}{5} \).

Understanding critical points is essential when solving an initial value problem because it allows us to analyze the behavior of solutions beyond just finding an explicit function. It aids in interpreting the context of a real-world scenario modeled by a differential equation by providing insights about the function's trends and extremities.