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# NDSolve & IVP DIFFeq HELP :(

Posted 10 years ago
 Hey guys!! doing my math homework and I cannot figure out how to use NDsolve...and When I looked at my friends graph of their behavior it looked completely different than the one I got so I deleted all of my work and now I turn to you math gurus to help me :(( I also had her explain to me the behavior of the solution over the interval and I pretended to understand what she said but I still don't understand question (c). Thanks for your help guys <3 Consider the initial-value problem (IVP) y'= cos(x) ? e^?y, with y(1) = 0 The Mathematica command DSolve fails with this problem (try it if you wish to see by yourself!), but there is the option to use the command NDSolve http://reference.wolfram.com/mathematica/ref/NDSolve.html, which uses special numerical algorithms to approximate the solution over an interval. (a) ?nd the solution y = f(x) with NDSolve, over the interval 0 ? x ? 2. (b) Graph f(x) on the interval 0 < x ? 2. Which part of the graph shows the initial condition? (c) Describe in words the behavior of the solution over the given interval. What is happening near x = 2? What happens with the computation of solution if the interval is extended to x = 3?
 You can use NDSolve like this ode = y'[x] == Cos[x] - Exp[-y[x]]; ic = y == 0; sol = First[y /. NDSolve[{ode, ic}, y, {x, 0, 2}]] Plot[sol[x], {x, 0, 2}] To see what happens as the range changes: Manipulate[ Module[{sol}, sol = First[y /. NDSolve[{ode, ic}, y, {x, 0, max}]]; Plot[sol[x], {x, 0, max}] ], {{max, 2, "max limit"}, .1, 4, .01, Appearance -> "Labeled"}, Initialization :> ( ode = y'[x] == Cos[x] - Exp[-y[x]]; ic = y == 0 ) ] You'll see the singularity after 2.0 show up. 