Question

\(\left(1+x^{2}\right) y^{\prime}-1 / 2 \cos ^{2} 2 y=0, y \rightarrow \frac{7}{2} \pi, x \rightarrow-\infty\).

Short answer

Question: Determine the general solution of the first-order differential equation \((1+x^2)y' - \frac{1}{2}\cos^2(2y) = 0\) with the asymptotic boundary condition \(y(x) \to \frac{7}{2}\pi, x \to -\infty\). Answer: The general solution of the given differential equation is \(\frac{1}{2}\tan(2y) = \frac{1}{2}\arctan(x) + C\).

Step-by-step solution

Identify the type of ODE and method

This ODE is non-linear and first order, so we will try to write it in the form of a separable ODE, which can be written as \(M(x)dx + N(y)dy = 0\). Then, we can integrate both sides to find the general solution.

Rewrite the ODE in the form of a separable ODE

The given differential equation is \((1+x^2)y' - \frac{1}{2}\cos^2(2y) = 0\). We have to rewrite it in the form of \(M(x)dx + N(y)dy = 0\). First, let's move the second term to the right side of the equation: \begin{align*} (1+x^2)y' = \frac{1}{2}\cos^2(2y) \end{align*} Now, divide both sides by \((1+x^2)\) and \(\cos^2(2y)\) to separate the variables x and y: \begin{align*} \frac{y'}{\cos^2(2y)} = \frac{1}{2(1+x^2)} \end{align*} Now, the equation is in the form of a separable ODE and can be written in terms of dx and dy as: \begin{align*} \frac{dy}{\cos^2(2y)} = \frac{1}{2(1+x^2)}dx \end{align*}

Integrate both sides

Now, we will integrate both sides with respect to their variables: \begin{align*} \int\frac{dy}{\cos^2(2y)} = \int\frac{1}{2(1+x^2)}dx \end{align*} The left side integral involves the substitution \(u = 2y\), which gives us \(du = 2dy\), leading to: \begin{align*} \frac{1}{2}\int\frac{du}{\cos^2(u)} = \frac{1}{2}\int\frac{dx}{1+x^2} \end{align*} Now, the left side integral is just the standard integral of the secant function squared, and the right side is the arctangent function: \begin{align*} \frac{1}{2}\tan(u) = \frac{1}{2}\arctan(x) + C \end{align*} Finally, change u back to y, giving the general solution: \begin{align*} \frac{1}{2}\tan(2y) = \frac{1}{2}\arctan(x) + C \end{align*}