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sql server reporting services barcode font Ordinary Differential Equations in Software
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1 dy 11.15 Solve the differential equation = 1 + y 2 = 0, y(0) = 1 , 0 x 1, using DSolve and NDSolve 2 dx and compare the results. SOLUTION
2 equation1 = DSolve y'[x] 1 + 1 y[x], y[0] 1 , y[x], x 2
Solve ifun : Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. y[x] 2 Tan 1 2 x + 2 ArcTan 1 2 2 solution1[x_] = equation1[[1, 1, 2]]; equation2 = NDSolve y'[x] 1 + 1 y[x]2, y[0] 1 ,[y[x],{x,0,1} 2 {{y[x] InterpolatingFunction[{{0., 1.}},<>][x]}} solution2[x_]= equation2[[1, 1, 2]]; tabledata = Table[{x, solution1[x], solution2[x]}, {x,0,1,.1}]; TableForm[tabledata, TableSpacing {1, 15}, TableHeadings {None, {"x","analytic", "numerical"}}] X 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1. analytic 1 1.15817 1.33582 1.53895 1.77601 2.05935 2.40786 2.85196 3.44406 4.28301 5.58016 numerical 1. 1.15817 1.33582 1.53895 1.77601 2.05935 2.40786 2.85196 3.44406 4.28301 5.58016 Ordinary Differential Equations
d 2 y dy dy 11.16 Plot the solution to the differential equation + +1 + y = 0, y(0) = 1, y '(0) = 0 for dt 2 dt dt 0 t 10. SOLUTION
2 solution1 = NDSolve[y''[t]+(y'[t]+ 1) y'[t] + y[t] 0,y[0] 1, y'[0] 0}, y[t],{t, 0, 10)] {{y[t] InterpolatingFunction[{{0.,10.}}, <>][t]}} Plot[y[t] /. solution,{t, 0, 10}, PlotRange All] 2 0.5 d2y dy 11.17 Plot the (five) solutions to + sin y = 0 for 0 x 30 using initial conditions y'(0) = 0, 2 + 0.3 dx dx y(0) = 2, 1, 0, 1, and 2. SOLUTION
Do[{solution = NDSolve[{y''[x]+ 0.3 y'[x]+ Sin[y[x]] 0, y[0] i, y'[0] 1}, y[x], {x, 0, 30}]; f[x_]= solution[[1,1,2]]; graph[i]= Plot[f[x], {x, 0, 30}, PlotStyle Hue[.2 i +.5], PlotRange All]}, {i, 2, 2}]; Show[graph[ 2], graph[ 1], graph[0], graph[1], graph[2]] When plotted in color, the five graphs are clearly distinguishable.
Ordinary Differential Equations
11.3 Laplace Transforms
In this section we describe an ingenious method for solving differential equations. Although the procedure can be used in a wide variety of problems, its real power lies in its ability to solve a differential equation whose right hand side is either discontinuous or zero except on a very short interval when its value is large. Because most of these types of problems arise within the context of time as the independent variable, we will express y and its derivatives as functions of t. We shall discuss Laplace transforms heuristically, and shall not concern ourselves with conditions sufficient for existence. If f is defined on the interval [0, ), the Laplace transform of f(t) is defined L { f (t )} = e st f (t ) dt
Its usefulness lies in the following properties, which we list without proof: L {1} = 1 L {sinh(bt)} = L {cosh(bt)} = b s2 b2 s s2 b2 b (s a) 2 b 2 s a (s a ) 2 b 2
L {t } = 1 2 s ! L {t n } = n+1 for positive integers n n s
L {e a t sinh(bt)} = L {e a t cosh(bt)} = L {e a t} = 1 L {sin(bt)} = s a b s2 + b2 s s2 + b2 b (s a) 2 + b 2 s a (s a) 2 + b 2
L { f '(t )} = s L { f (t )} f (0) L { f ''(t )} = s 2 L { f (t )} sf (0) f '(0) L {af (t ) + bg(t )} = a L If F (s) = L
L {cos(bt)} = L {e a t sin(bt)} = L {e a t cos(bt)} = { f (t )} + b L {g(t )} { f (t )} , then L {ea t f (t )} = F (s a) Mathematica computes the Laplace transform of a function, f, by the invocation of the command LaplaceTransform. LaplaceTransform[f[var1], var1, var2] computes the Laplace transform of the function f, with independent variable var1, and expresses it as a function of var2.

