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Use the backward Euler method?

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1 Answer

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.