Question

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

Short answer

Question: Determine the minimum value of the solution to the initial value problem: \(y' = 2(1+x)(1+y^2)\), \(y(0) = 0\). Answer: The minimum value of the solution is \(y(-1) = \tan(-1)\) at \(x = -1\).

Step-by-step solution

Separate variables

We can rewrite the given differential equation: $$ y'=2(1+x)(1+y^2) $$ as $$ \frac{dy}{dx} = 2(1+x)(1+y^2) \Rightarrow \frac{1}{1+y^2} dy = 2(1+x) dx. $$ Now the variables are separated.

Integrate both sides

Next, we will integrate both sides of the equation with respect to their respective variables: $$ \int \frac{1}{1+y^2} dy = \int 2(1+x) dx. $$ The antiderivatives are: $$ \arctan(y) = x^2 + 2x + C. $$ Now we can apply the initial condition, y(0) = 0, to find the constant C: $$ \arctan(0) = C \Rightarrow C = 0. $$ So the equation becomes: $$ \arctan(y) = x^2 + 2x. $$

Solve for y

To find the explicit solution for y(x), we will apply the inverse tangent function: $$ y(x) = \tan(x^2 + 2x). $$

Find the minimum value

To find where the solution attains a minimum value, we must find the critical points (where the derivative is equal to 0 or undefined) of y(x). Using the chain rule, we find the derivative: $$ y'(x) = \frac{d}{dx} \tan(x^2 + 2x) = \sec^2(x^2 + 2x) \cdot (2x + 2), $$ where the derivative is undefined when \(\sec^2(x^2 + 2x)\) is infinity, which corresponds to \(\cos(x^2 + 2x) = 0\). To find where the derivative is 0, we set the expression inside the parenthesis to 0: $$ 2x + 2 = 0 \Rightarrow x = -1. $$ Additionally, the expression \(\sec^2(x^2 + 2x)\) is always non-negative, so it cannot cancel out the term (2x + 2). Now, we evaluate y(x) at x = -1: $$ y(-1) = \tan((-1)^2 + 2(-1)) = \tan(1-2) = \tan(-1). $$ Thus, the solution attains its minimum value at x = -1, and the minimum value is y(-1) = \(\tan(-1)\).

Key concepts

Initial Value Problem

In differential equations, an initial value problem involves finding a specific solution that not only satisfies the differential equation but also meets given initial conditions. Initial conditions are values specified for the solution at a particular point, and in our problem, the condition is that when \( x = 0 \), \( y = 0 \).

This condition is crucial because it allows us to determine the particular solution among the family of solutions that the differential equation could have. It enables solving for any arbitrary constants that result from integration. The integration step will usually yield a constant, represented as \( C \), which can be resolved by substituting the initial condition into the integrated equation.

In our example, we found that \( C = 0 \) by using the condition \( \arctan(0) = 0 \), making the particular solution suitable for our problem.

Variable Separation

The method of variable separation is a technique used to solve differential equations by rewriting the equation in such a way that each variable appears on different sides of the equation. This involves moving terms involving \( y \) to one side and terms involving \( x \) to the other.

For the given differential equation, we had \( y' = 2(1+x)(1+y^2) \), which we rearranged as \( \frac{1}{1+y^2} dy = 2(1+x) dx \). This successful separation of variables allowed us to integrate both sides with respect to their respective variables later on.

Variable separation is essential because it permits us to solve by integration, reducing the complexity of finding a solution to a simpler problem of computing integrals.

Critical Points

The critical points of a function are where its derivative is zero or undefined. In solving differential equations, locating these points helps determine where the function achieves local maxima, minima, or saddle points. This is important for analyzing the behavior of solutions.

After finding the explicit expression for \( y(x) \), we derived its derivative to find the critical points. The derivative \( y'(x) = \sec^2(x^2 + 2x) \cdot (2x + 2) \) becomes zero when \( 2x + 2 = 0 \), giving us \( x = -1 \).

Evaluating \( y(x) \) at \( x = -1 \) permitted us to identify the minimum value of the solution. Recognizing such critical points can reveal where a function achieves specific extremum values.

Inverse Functions

Inverse functions are functions that reverse the effect of the original function, taking an output back to its input. In the context of our differential equation problem, we encountered the need for an inverse function while solving \( \arctan(y) = x^2 + 2x \) to find \( y \).

We used the inverse tangent function (the "arctan") to isolate \( y \), leading us to \( y = \tan(x^2 + 2x) \). Employing inverse functions is often a necessary step in expressing one variable explicitly in terms of another after certain transformations have been applied.

The ability to identify and apply the correct inverse functions is fundamental in solving equations, especially when aiming to express the solution in a more useful form for interpretation or further analysis.