Blowup in finite time Consider the initial value problem y'(t) = yⁿ + 1, y(0) = y₀, where n is a positive integer.
b. Solve the initial value problem with n = 2 and y₀ = 1/√2.

Verified step by step guidance

1. Start with the given initial value problem: \(y'(t) = y^2 + 1\) with initial condition \(y(0) = \frac{1}{\sqrt{2}}\).
2. Rewrite the differential equation in separable form: \(\frac{dy}{dt} = y^2 + 1\) implies \(\frac{dy}{y^2 + 1} = dt\).
3. Integrate both sides: integrate \(\int \frac{dy}{y^2 + 1}\) on the left and \(\int dt\) on the right. Recall that \(\int \frac{dy}{y^2 + 1} = \arctan(y) + C\).
4. After integration, write the implicit solution: \(\arctan(y) = t + C\). Use the initial condition \(y(0) = \frac{1}{\sqrt{2}}\) to solve for the constant \(C\) by substituting \(t=0\) and \(y=\frac{1}{\sqrt{2}}\).
5. Finally, solve for \(y\) explicitly by taking the tangent of both sides: \(y = \tan(t + C)\), which gives the solution to the initial value problem.

Key Concepts

Separable Differential Equations

A separable differential equation can be written as dy/dt = g(y)h(t), allowing variables y and t to be separated on opposite sides of the equation. This enables integration with respect to each variable independently, which is essential for solving the given initial value problem.

Initial Value Problem (IVP)

An initial value problem specifies a differential equation along with a condition y(t₀) = y₀. This condition allows determination of the particular solution from the family of general solutions by solving for the integration constant.

Finite Time Blowup

Finite time blowup occurs when the solution to a differential equation becomes unbounded in a finite time interval. For nonlinear equations like y' = y² + 1, solutions can grow rapidly and approach infinity at a finite time, which is important to analyze after solving the IVP.