Final Answer:
The solution to the Initial Value Problem y' = tan(y), y(ln(6)) = 2π, is y = π.
Explanation:
The given initial value problem is a first-order ordinary differential equation (ODE) with a separable variable. We can start by separating the variables to solve it. Begin by rearranging the equation: dy / dx = tan(y). Separate the variables by bringing all y terms to the left-hand side and x terms to the right-hand side: dy / tan(y) = dx.
Now, integrate both sides individually. The integral of dy / tan(y) involves integrating the tangent function, which results in -ln|cos(y)| = x + C, where C is the constant of integration. Simplify further by solving for y, taking the exponential of both sides: |cos(y)| = e^(-x-C). However, the absolute value complicates the solution.
The initial condition y(ln(6)) = 2π allows us to find the specific solution. When x = ln(6), the equation becomes |cos(y)| = e^(-ln(6)-C), which simplifies to |cos(y)| = 1 / 6 * e^(-C). This condition restricts the solution to cos(y) = ±1 / 6 * e^(-C).
The given initial condition 2π for y(ln(6)) implies that cos(y) = 1 / 6 * e^(-C). Since y = 2π satisfies this condition, the solution is y = π, as it's the only value that satisfies the given initial condition for y(ln(6)) = 2π.