Question

Find a solution of the initial-value problem.
y' = −(1/4)y² y(0) = 0.25
y =

Answer 1

The solution to the initial-value problem y' = −(1/4)y^2 with y(0) = 0.25 is found by separating variables, integrating both sides, applying the initial condition to find the integration constant, and then expressing y as a function of t, resulting in y = -1/(t+16).

Step-by-step explanation:

The student is tasked with solving an initial-value problem with the differential equation y' = −(1/4)y^2 and the initial condition y(0) = 0.25. This is a separable differential equation, and we can integrate both sides to find the solution.

1. Separate the variables: dy/y^2 = -1/4 dt.
2. Integrate both sides: ∫dy/y^2 = -1/4 ∫dt, which gives -1/y = -1/4 t + C.
3. Using the initial condition y(0) = 0.25, find C: -1/0.25 = -1/4 (0) + C, hence C = -4.
4. Write the solution: -1/y = -1/4 t - 4, or after rearranging y = -1/(t+16).

This gives us the solution to the initial-value problem, where y is a function of t.