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I am not sure why you do not want to use DSolve but it works perfectly in your case. f[t_] := Piecewise[ {{t^2 , 0 <= t < 1}, { 1, t > 1}}] DSolve[ {y''[t] - y[t] == f[t] , y[0] == 0, y'[0] == 0 }, y[t], t] gives correct solution of your equation. If you insist using Laplace transform in this problem, this can be done as follows, lap = ...


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In Mathematica, however, the search is simple, for instance: GraphData[{"Bicubic", "Asymmetric", "Planar"}] (* {{"Cubic", {24, 3}}, {"Cubic", {54, 1}}, {"CubicPolyhedral", 9}, {"CubicPolyhedral", 18}, {"CubicPolyhedral", 42}, {"CubicPolyhedral", 50}, "GreatRhombicosidodecahedralGraph", \ "GreatRhombicuboctahedralGraph", {"Prism", 6}, {"Prism", ...



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