sql server reporting services barcode font Plot the vector field F(x, y) = y i + x j . in Software

Printer QR Code in Software Plot the vector field F(x, y) = y i + x j .

EXAMPLE 9 Plot the vector field F(x, y) = y i + x j .
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VectorPlot [{ y, x}, {x, 5, 5}, {y, 5, 5}]
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Any first-order differential equation can be used to define a vector field. Indeed, the vector field i + f(x, y) j, corresponding to the equation dy = f ( x , y) , generates a field whose vectors are tangent to the solution. The
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next example, although simple, illustrates this nicely.
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Ordinary Differential Equations
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EXAMPLE 10 Plot the vector field of the solution of the equation dy = 2 x . The solutions to this equation, parabolas dx y = x2 + c, can be seen quite vividly.
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VectorPlot[{1, 2 x}, {x, 1, 1}, {y, 1, 1}, Axes Automatic]
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EXAMPLE 11 In this example we plot the vector field generated by the equation
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dy = 2 x + y. Then the soludx
tions with initial conditions y(0) = 2, 1, 0, 1, and 2 are plotted on the vector field for comparison.
vf = VectorPlot[{1, 2 x + y}, {x, 2, 1}, {y, 4, 6}, Axes Automatic, VectorScale Small]
4 2.0 1.5 1.0 0.5 0.0 0.5 1.0
Ordinary Differential Equations
solutions = Table[DSolve[{y'[x] 2 x + y[x], y[0] k}, y[x], x], {k, 2, 2}]; Do[g[k]= Plot[solutions[[k, 1, 1, 2]], {x, 2, 1}, PlotRange All, Frame True, PlotStyle Thickness[.005]], {k, 1, 5}] Show[g[1], g[2], g[3], g[4], g[5], vf, AspectRatio 1, PlotRange { 4,6}]
4 2.0
1.5
1.0
0.5
A system of differential equations consists of n differential equations involving n + 1 variables. Solving a system of differential equations with Mathematica is similar to solving a single equation.
This example illustrates how to solve the system dx = t 2 , dy = t 3 with initial conditions dt dt x(0) = 2, y(0) = 3. The equation and its initial conditions are contained within a list.
EXAMPLE 12
solution = DSolve[{x'[t] t2, y'[t] t3, x[0] 2, y[0] 3}, {x[t], y[t]}, t] x[t] 1 (6 + t3 ), y[t] 1 (12 + t4 ) 3 4
Instead of specifying the values of f and its derivatives at a single point, values at two distinct points may be given. The problem of solving the differential equation then becomes known as a boundary value problem. However, unlike initial value problems, which can be shown to have unique solutions for a wide variety of cases, a boundary value problem may have no solution even for the simplest of equations.
EXAMPLE 13 The equation
d2y + y = 0 with boundary conditions y(0) = 0, y( ) = 1 has no solution. dx 2 DSolve[{y''[x]+ y[x] 0, y[0] 0, y[o] 1}, y[x], x]
DSolve bvnul : For some branches of the general solution, the given boundary conditions lead to an empty solution.
{} The same equation with y(0) = 0, y( /2) = 1 has a unique solution. DSolve[{y''[x]+ y[x] 0, y[0] 0, y[o/2] 1}, y[x], x]
{{y[x]
Sin[x]}}
Ordinary Differential Equations
SOLVED PROBLEMS
11.1 Solve the differential equation dy = x y with initial condition y(1) = 2 and graph the solution for 2 x 2.
SOLUTION
solution = DSolve[{y'[x] x y[x], y[1] 2}, y[x], x]
{{y[x] 2
1 + x2 2 2
8 6 4 2
Plot[y[x] /.solution, {x, 2, 2}]
11.2 Plot the vector field for the differential equation of Problem 11.1.
SOLUTION
VectorPlot[{1, x y}, {x, 2, 2}, {y, 10, 10}, VectorScale Tiny, Axes Automatic]
10 2 1 0 1 2
11.3 Plot the vector field for the equation dy = x 2 + y together with its solutions for y(0) = 0, 1, 2, 3, and 4.
SOLUTION
vf = VectorPlot [{1, x2 + y}, {x, 0, 1}, {y, 0, 12}, Axes Automatic, VectorScale Tiny];
Ordinary Differential Equations
solutions = Table[DSolve[{y'[x] x2 + y[x], y[0] k}, y[x], x], {k, 0, 4}]; Do[g[k]= Plot[solutions[[k, 1, 1, 2]], {x, 0, 1}, PlotStyle Thickness[.007], PlotRange All], {k, 1, 5}] Show[g[1], g[2], g[3], g[4], g[5], vf, Frame True, AspectRatio 1]
0 0.0 0.2 0.4 0.6 0.8 1.0
11.4 The escape velocity is the minimum velocity with which an object must be propelled in order to escape the gravitational field of a celestial body. Compute the escape velocity for the planet Earth.
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