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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
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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.