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First-order linear homogeneous differential

This non-linear homogeneous differential equation of first order cannot be solved analytically. Therefore a numerical approach is used. A simple integration scheme is... [Pg.77]

Equations (2) and (3) represent two coupled, first order, linear, non-homogenous differential equations of the form dxidt + a(> x =. For radical pairs in liquid solution x and f(x) are... [Pg.430]

The mathematical problem posed is the solution of the simultaneous differential equations which arise from the mass-action treatment of the chemistry. For the homogeneous, well-mixed reactor, this becomes a set of ordinary, non-linear, first-order differential equations. For systems that are not... [Pg.120]

We need the tangential components of E and H (namely Ex, Ey, Hx, and Hy) as explicit variables, so we want to get rid of Hz and Ez this is done by solving Eqs. (2.15.14) and (2.15.15) explicitly and simultaneously for Ez and Hz in terms of Ex, Ey, Hx, and Hy. These expressions for Ez and Hz are then substituted into the remaining four differential equations, to produce four linear homogeneous first-order differential equations in the four tangential field variables Ex, Ey, Hx, Hz. For convenience, we define a 4 x 1 generalized field vector dr. [Pg.93]

Since a and 3 are represented by 4 x 4 matrices, the wave function / must also be a four-component function and the Dirac wave equation (3.9) is actually equivalent to four simultaneous first-order partial differential equations which are linear and homogeneous in the four components of P. According to the Pauli spin theory, introduced in the previous chapter, the spin of the electron requires the wave function to have only two components. We shall see in the next section that the wave equation (3.9) actually has two solutions corresponding to states of positive energy, and two corresponding to states of negative energy. The two solutions in each case correspond to the spin components. [Pg.76]

The radiation balance of a layer with the thickness d having an infinitely large surface, irradiated homogeneously from one side with exciting radiation, is given by the solution of four coupled linear first-order differential equations (Eqs. 3.5-1...4). This is a boundary value problem, with the definitions given in Fig. 3.5-2. We are discussing... [Pg.139]

Remember 2.1 The general solution to nonhomogeneous linear first-order differential equations can be obtained as the product of function to be determined and the solution to the homogeneous equation. [Pg.27]

The time dependent behavior of the flux is determined by solving a system of six linear non-homogeneous differential equations of the first order, with constant coefficients, involving Ty(t) and Hiv t). When the initial conditions are fulfilled the expression for the flux takes the form... [Pg.273]

We present a brief introduction to coupled transport processes described macroscopically by hydrodynamic equations, the Navier-Stokes equations [4]. These are difficult, highly non-linear coupled partial differential equations they are frequently approximated. One such approximation consists of the Lorenz equations [5,6], which are obtained from the Navier-Stokes equations by Fourier transform of the spatial variables in those equations, retention of first order Fourier modes and restriction to small deviations from a bifurcation of an homogeneous motionless stationary state (a conductive state) to an inhomogeneous convective state in Rayleigh-Benard convection (see the next paragraph). The Lorenz equations have been applied successfully in various fields ranging from meteorology to laser physics. [Pg.83]

The problem is separable for a bare homogeneous reactor. However, only the case of a step input of reactivity, i.e., the case of a constant value of p, is easily solved. In this case, the kinetic equations are readily reduced to a second order (for the case of one delayed neutron group) homogeneous linear differential equation with constant coefficients. For an input of positive reactivity two solutions arise, of the form and where o>i > 0 and 0)2 < 0. The first solution controls the persisting exponential rise of the flux, where it is recalled that T = l/o>i is the reactor period, and the second solution which rapidly becomes small is called the transient solution. [Pg.265]


See other pages where First-order linear homogeneous differential is mentioned: [Pg.471]    [Pg.471]    [Pg.21]    [Pg.456]    [Pg.318]    [Pg.180]    [Pg.318]    [Pg.288]    [Pg.283]    [Pg.460]    [Pg.318]    [Pg.460]    [Pg.21]    [Pg.275]    [Pg.212]    [Pg.260]    [Pg.94]    [Pg.94]    [Pg.482]    [Pg.62]   


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