Question

Use the methods of this section to solve the following differential equations. Compare computer solutions and reconcile differences. $$y^{\prime}=\cos (x+y) \quad \text { Hint: } \quad \text { Let } u=x+y ; \text { then } u^{\prime}=1+y^{\prime}$$

Short answer

The solution to the differential equation is: \( y = 2 \arctan(\frac{x+C}{2}) = x \).

Step-by-step solution

Identify Substitution

Given the hint, let us denote the substitution: \( u = x + y \).

Calculate Derivative

Compute the derivative of \( u \) with respect to \( x \): \( u' = \frac{du}{dx} = 1 + \frac{dy}{dx} = 1 + y' \).

Substitute Differential Equation

Substitute the given differential equation \( y' = \cos(x + y) \) into the derivative of \( u \): \( u' = 1 + \cos(x + y) = 1 + \cos(u) \).

Solve the Differential Equation

Now solve the differential equation \( \frac{du}{dx} = 1 + \cos(u) \). Separate the variables to obtain: \( \frac{du}{1 + \cos(u)} = dx \). Integrate both sides.

Integrate Both Sides

Integrate: \( \int \frac{du}{1 + \cos(u)} \). Using the identity \( 1 + \cos(u) = 2 \cos^2(\frac{u}{2}) \), rewrite the integral: \( \int \frac{du}{2 \cos^2(\frac{u}{2})} = \int \sec^2(\frac{u}{2})d(\frac{u}{2}) = 2 \int \sec^2(\frac{u}{2})d(\frac{u}{2}) \). This simplifies to \( 2 \tan(\frac{u}{2}) \).

Integrate Right Side

Integrate \( dx \) on the right side: \( \int dx = x + C \).

Combine the Results

Set the results of both integrals equal: \( 2 \tan(\frac{u}{2}) = x + C \).

Solve for y

Since \( u = x + y \), we have: \( 2 \tan(\frac{x + y}{2}) = x + C \). To solve for \( y \), rearrange the equation: \( y = 2 \arctan(\frac{x+C}{2}) = x \).

Key concepts

separation of variables

Separation of variables is a technique often used to solve differential equations, where we express the equation in a form where each variable can be integrated separately on different sides of the equation. In our exercise, once we substituted the given equation using a proper substitution method, we separated the variables so that we could integrate both sides easily. For example, separating the variables gave us one side with the integrand involving only the new variable (like u) and the other side involving only x.

substitution method

The substitution method involves replacing a variable in the differential equation with another variable to simplify the equation. Here, we used the hint provided in the exercise to replace x + y with u. This transformed our original problematic differential equation into a simpler form. Using this new variable, derivatives were recalculated, making the equation easier to solve. For instance, labeling u as x + y allowed us to then find u' which in turn simplified our differential equation to one involving u and its derivative.

integration techniques

Integration techniques are essential for solving differential equations after separating the variables. These techniques involve various known integration rules and identities. For instance, in the provided solution, we integrated both sides after simplifying the variables. We utilized special trigonometric identities to transform the integral into a form we could handle. This was demonstrated when we converted the integral of 1 / (1 + cos(u)) by employing the known identity involving secant squared.

trigonometric identities

Trigonometric identities were crucial in simplifying our integral. One key identity used in this problem was 1 + cos(u) = 2 cos^2(u/2). Trigonometric identities like this help in transforming integrals into a more solvable form. When we rewrote our integral using this identity, it became easier to handle, and we could directly use the known integral of secant squared. Understanding and applying these identities make integrals involving trigonometric functions much easier to calculate.

initial value problems

In many cases, differential equations come with initial conditions. These initial conditions help pinpoint a specific solution from a family of possible solutions. In initial value problems, after solving the differential equation, we substitute the initial condition values into our general solution to determine the constants of integration. While this particular exercise didn't specify an initial condition, if it had, we would have used it to find the exact value of the constant, C, ensuring our solution fits the provided initial value exactly.