Use the backward Euler method Use the backward Euler method

Use the backward Euler method?
Use the backward Euler method with h=0.025 of y\'=(t^2-y^2)sin y, y(0)=-1 at t=0.1, 0.2, 0.3, and 0.4.

Answer: -0.926341, -0.867163, -0.820279, -0.784275

Solution

For backward Euler method,
y(n+1) = y\'(n+1) h + y(n), where y\'(n+1) = (t(n+1)^2 - y(n+1)^2)sin(y(n+1))
Since y(n+1) can not be solved explicitly, you probably have to use some numerical ways, such as Newton\'s method, to solve y(n+1) for each step.