Question

Determine whether the following equations are separable. If so, solve the initial value problem. $$y^{\prime}(t)=\cos ^{2} y, y(1)=\frac{\pi}{4}$$

Short answer

Question: Solve the initial value problem for the given differential equation and initial condition. Differential Equation: $$ y'(t) = \cos^2{y} $$ Initial Condition: $$ y(1) = \frac{\pi}{4} $$ Answer: The solution to the initial value problem is $$ y(t) = \arctan{t} $$ with the initial condition $$ y(1) = \frac{\pi}{4} $$.

Step-by-step solution

Determine if the equation is separable

The equation is given as: $$ y'(t) = \cos^2{y} $$ Compare it to the form \(y'(t) = f(t)g(y)\). We can rewrite the right-hand side as: $$ y'(t) = (1)(\cos^2{y}) $$ Clearly, \(f(t) = 1\) and \(g(y) = \cos^2{y}\). The equation is separable.

Apply separation of variables

Now that we have confirmed that the equation is separable, we will separate the variables and rewrite the equation in the form: $$ \frac{dy}{\cos^2{y}} = dt $$

Integrate both sides

Integrate both sides of the equation: $$ \int \frac{dy}{\cos^2{y}} = \int dt $$ The left side can be integrated using the substitution technique with \(u = \tan{y}\) and \(du = \sec^2{y}dy\), giving: $$ \int du = \int dt $$

Solve for y(t)

Integrate both sides: $$ u = t + C $$ Substitute back for \(y\): $$ \tan{y} = t + C $$ So, we have $$ y(t) = \arctan{(t + C)} $$

Apply the initial condition

Now, we'll use the initial condition \(y(1) = \frac{\pi}{4}\) to find the constant \(C\). $$ y(1) = \arctan{(1 + C)} = \frac{\pi}{4} $$ We can find the value of \(C\) by taking the tangent of both sides: $$ \tan{y(1)} = 1 + C \Rightarrow \tan{\frac{\pi}{4}} = 1 + C $$ Since \(\tan{\frac{\pi}{4}} = 1\), we have: $$ 1 = 1 + C \Rightarrow C = 0 $$

Write the final solution

Now, substitute the value of \(C\) back into the equation for \(y(t)\): $$ y(t) = \arctan{(t + 0)} = \arctan{t} $$ This is the solution to the initial value problem: $$ y(t) = \arctan{t}, y(1) = \frac{\pi}{4} $$

Key concepts

Initial Value Problem

An initial value problem is a type of differential equation along with a specified condition called the initial condition. This initial condition usually involves the value of the unknown function at a certain point. In simple terms, it tells us where the solution to the differential equation should begin.

To capture this, an initial value problem is often framed as follows:

• A differential equation in terms of the function and its derivatives.
• An initial condition that specifies the value of the function at a particular point.

In the exercise above, the differential equation is given by:\[ y'(t) = \cos^2{y} \]and the initial condition is:\[ y(1) = \frac{\pi}{4} \]This means we are looking for a function \( y(t) \) that satisfies both the equation and the condition at \( t = 1 \). The initial condition helps to find a unique solution from the family of possible solutions, ensuring that the solution fits the problem context.

Integration Techniques

To solve separable differential equations, integration techniques are essential. Once we have separated the variables, each side of the equation is integrated separately to help find the solution.

For instance, in the given problem, after separating variables, we arrived at:\[ \int \frac{dy}{\cos^2{y}} = \int dt \]The left side of the equation often requires specific integration techniques, such as substitution, while the right side is typically straightforward to integrate, usually resulting in just the variable, \( t \), plus a constant.

Understanding which integration technique to apply depends on the form of the functions involved:

• Direct Integration: Used when simple antiderivatives are involved.
• Substitution: Helpful for transforming a difficult integral into a simpler one, often used when a function's derivative is present in the equation.

By properly applying these techniques, we can evaluate integrals and proceed to solve the equation.

Substitution Method

The substitution method is a powerful tool in integration, often used for simplifying the problem or when part of the integral matches a derivative of a function. This method involves substituting part of the original integral with a new variable, making it easier to integrate.

In the problem given, we used substitution to integrate \( \frac{dy}{\cos^2{y}} \). By letting \( u = \tan{y} \), the derivative \( du = \sec^2{y}dy \) was used. Since \( \frac{dy}{\cos^2{y}} = \tan{y} \), this substitution simplified the equation to:\[ \int du = \int dt \]

• First, identify a substitution that simplifies the integral.
• Make the substitution, replacing the original variable with the new one.
• Solve the resulting simpler integral.

Finally, it is important to remember to substitute back the original variable once the integration is complete, resulting in a solution in terms of the original variables used in the problem.