- Replace dy/dt with (y(t+1)-y(t))/dt
- Simplify it so that y(t+1) is on LHS and rest all on RHS. You will get a equation, RHS has all known values and LHS is the value which we seek.
- Initliaze the required arrays and first value of y i.e y = 1 at t=0;
- Create time array with specified time step. t = 0:0.001:0.25.
- Run a loop using time.
Apply the finite difference
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How to apply the finite difference to the following equation of y(t); 𝑑𝑦/𝑑𝑡 = 2𝑦^3 + 5. Given at t=0, y=1. Take time steps at Δt=0.001 and step the solution forward to t=0.25 ?
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KSSV
2020 年 7 月 14 日
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