Problem 36 - Calculus Single Variable

Question

Find the solution of the given initial value problem. $$ \frac{d y}{d x}=2 x \cos (y(x)) \quad y(0)=0 $$

Step-by-Step Solution

Answer

The solution is \( \sec(y) + \tan(y) = e^{x^2} \).

Step 1: Identify the Type of Differential Equation

The given differential equation is \( \frac{dy}{dx} = 2x \cos(y) \), which is a first-order separable differential equation. This means we can separate variables to solve it.

Step 2: Separate Variables

Rearrange the equation to separate the variables. Move all terms involving \( y \) to one side and \( x \) to the other:\[ \frac{dy}{\cos(y)} = 2x \, dx \]

Step 3: Integrate Both Sides

Integrate both sides of the equation. The left side will be integrated with respect to \( y \) and the right side with respect to \( x \):\[ \int \frac{dy}{\cos(y)} = \int 2x \, dx \]

Step 4: Find the Integral of Each Side

Calculate each integral:- The left side becomes \( \int \sec(y) \, dy = \ln |\sec(y) + \tan(y)| \)- The right side becomes \( x^2 + C \) where \( C \) is a constant of integration. Thus, \( \ln|\sec(y) + \tan(y)| = x^2 + C \).

Step 5: Apply the Initial Condition

Given the initial condition \( y(0) = 0 \), substitute \( x = 0 \) and \( y = 0 \) into the equation from Step 4:\[ \ln|\sec(0) + \tan(0)| = 0^2 + C \]Since \( \sec(0) = 1 \) and \( \tan(0) = 0 \), we have:\[ \ln|1 + 0| = C \Rightarrow C = 0 \]

Step 6: Simplify the Expression

With \( C = 0 \), the equation becomes:\[ \ln|\sec(y) + \tan(y)| = x^2 \]Solve for \( y \) by exponentiating both sides:\[ |\sec(y) + \tan(y)| = e^{x^2} \]Since \( \sec(y) + \tan(y) \) must be positive:\[ \sec(y) + \tan(y) = e^{x^2} \]

Step 7: Solve for y

To express \( y \) explicitly could be complex, but the equation \( \sec(y) + \tan(y) = e^{x^2} \) is our solution implicitly. This involves inverse trigonometric functions, which typically requires numerical or graphical techniques for exact solutions.

Key Concepts

Separation of variables

The process of 'Separation of Variables' is a technique used to solve differential equations, particularly when dealing with first-order separable differential equations. The main goal here is to rearrange the equation so that each variable and its derivatives are on opposite sides of the equation. This allows separate integration of each side. For example, given the differential equation \( \frac{dy}{dx} = 2x \cos(y) \), we move all terms involving \( y \) to one side and terms involving \( x \) to the other, resulting in the expression:

• \( \frac{dy}{\cos(y)} = 2x \, dx \)

This rearrangement makes it possible to integrate both sides separately, setting the stage for solving the equation. This step is crucial because it simplifies the equation into two distinct integral problems. Each side will then be solved independently by integration, which will help draw closer to the function that solves the original differential equation.

Initial value problem

An 'Initial Value Problem' (IVP) is a type of differential equation problem that includes conditions at a start point, known as initial conditions. These conditions allow us to find a specific solution among the infinite general solutions that a differential equation might have. The given initial condition in our problem is \( y(0) = 0 \). This means when \( x = 0 \), \( y \) should equal zero. Applying this initial condition helps determine the constant of integration. For example, in our step-by-step solution when integrating, we get to a point where \( \ln|\sec(y) + \tan(y)| = x^2 + C \). By substituting the initial condition \( x = 0 \) and \( y = 0 \), we derive the constant \( C \).

• \( \ln|\sec(0) + \tan(0)| = 0^2 + C \)
• \( 1 = C \)

In this case, since \( \ln|1| = 0 \), it shows that \( C = 0 \). This application of the initial condition ensures that the solution fits the specific scenario given, rather than just providing a general formula.

Integration

'Integration' is a fundamental mathematical process used to solve differential equations once the variables have been separated. Integration computes the antiderivative, helping undo the differentiation applied to a function. When addressing an equation like \( \frac{dy}{dx} = 2x \cos(y) \), after separation, you will integrate both sides: \[ \int \frac{dy}{\cos(y)} = \int 2x \, dx \]Integrating these terms individually results in:

• The left integral: \( \int \sec(y) \, dy = \ln |\sec(y) + \tan(y)| \)
• The right integral: \( \int 2x \, dx = x^2 + C \)

Integration allows us to revert back from the derivatives form of functions to the functions themselves. The addition of the constant \( C \) arises because indefinite integrals represent a family of functions, and the exact value of \( C \) is determined with the help of initial conditions.