Question

Find the solution to the initial-value problem.\(y^{\prime}=-2 x \tan (y), y(0)=\frac{\pi}{2}\)

Short answer

The solution is \( y = \arcsin(e^{-x^2}) \).

Step-by-step solution

Separate Variables

To solve the differential equation, we start by separating the variables. The given equation is \( y' = -2x \tan(y) \). Rewriting in terms of \( dy \) and \( dx \):\[\frac{dy}{\tan(y)} = -2x \, dx\]

Integrate Both Sides

Next, integrate both sides to solve for \( y \).For the left side, integrate \( \int \frac{dy}{\tan(y)} \), which can be rewritten as \( \int \cot(y) \, dy \).For the right side, integrate \( \int (-2x) \, dx \).The integrations lead to:\[\ln|\sin(y)| = -x^2 + C\]

Solve for the Integration Constant

Use the initial condition \( y(0) = \frac{\pi}{2} \) to determine the integration constant \( C \).Plugging in the initial values:\[\ln|\sin(\frac{\pi}{2})| = -0^2 + C\]Since \(|\sin(\frac{\pi}{2})| = 1\),\[\ln(1) = C \]So, \( C = 0 \).

Express the Solution

Substitute \( C = 0 \) back into the integrated equation.\[\ln|\sin(y)| = -x^2\]Exponentiating both sides to solve for \( y \) yields:\[|\sin(y)| = e^{-x^2}\]Thus, \( \sin(y) = \pm e^{-x^2} \). Choose \( \sin(y) = e^{-x^2} \) consistent with the initial condition \( y(0) = \frac{\pi}{2} \), ensuring \( \sin(y(0)) = 1 \).

Final Answer

The solution to the initial value problem is:\[ y = \arcsin(e^{-x^2}) \]

Key concepts

Understanding Initial Value Problems

An initial value problem in the context of differential equations is a problem where you're given a differential equation and an initial condition. This initial condition helps determine the specific solution from a family of possible solutions. In our example, the differential equation is given by\(y^{\prime}=-2 x \tan(y)\).
The initial condition is \(y(0)=\frac{\pi}{2}\).
This means that when \(x = 0\), \(y\) must equal \(\frac{\pi}{2}\).Why Initial Conditions Matter- Initial conditions provide a specific point that the solution curve must pass through.- They help in finding a unique solution for differential equations that could otherwise represent many possible outcomes.- Don’t confuse differential equations with indefinite integration, where a constant of integration always remains undefined unless an initial condition like \(y(0)=\frac{\pi}{2}\) is given. It pinpoints the exact value of the integration constant.

Separation of Variables in Differential Equations

Separation of variables is a method used to solve differential equations, where the equation can be rewritten to isolate each variable and its derivatives. For our differential equation, \( y' = -2x \tan(y) \), the method involves rewriting this equation in terms of \(dy\) and \(dx\), separating them to each side.In practice:- Arrange terms to allow each variable to have its own side of the equation: \[\frac{dy}{\tan(y)} = -2x \, dx\].- By rewriting the equation, we prepare it for integration, which is the next step.Steps in Separation of Variables- Rearrange terms to isolate variables.- Integrate both sides to progress towards solving the equation.- This method often involves algebraic manipulation to get appropriately separated forms.Separation of variables works well when the equation can be easily arranged in this format, and it simplifies solving these types of differential equations.

Role of the Integration Constant

When integrating both sides of an equation in a separation of variables problem, an integration constant \(C\) appears. This constant \(C\) is crucial as it represents an infinite number of solutions in a family of curves generated by the differential equation.In the step-by-step solution, integration on both sides led to the equation \(\ln|\sin(y)| = -x^2 + C\). Here, \(C\) was determined using the initial condition, \(y(0) = \frac{\pi}{2}\):- Substitute \(x = 0\) and \(y = \frac{\pi}{2}\) into the integrated equation.- Calculate to find \(C\), which was found to be zero as \(|\sin(\frac{\pi}{2})| = 1\) implies \(\ln(1) = 0\).Understanding the Integration Constant- Without \(C\), the solution would not satisfy the initial condition.- Determining \(C\) with given conditions ensures the solution is specific and exact.- It emphasizes that every integral without limits represents a family of curves, and the constant makes them unique for a specific case.