نتایج جستجو برای: ordinary differential equations
تعداد نتایج: 499181 فیلتر نتایج به سال:
where z is a new variable. This exemplifies the procedure for an arbitrary ODE. The usual choice for the new variables is to let them be just derivatives of each other (and of the original variable). Occasionally, it is useful to incorporate into their definition some other factors in the equation, or some powers of the independent variable, for the purpose of mitigating singular behavior that ...
Effect of nonlinear thermal radiation on the unsteady magnetohydrodynamic slip flow of Casson fluid between parallel disks in the presence of thermophoresis and Brownian motion effects are investigated numerically. A similarity transformation is employed to reduce the governing partial differential equations into ordinary differential equations. Further, Runge-Kutta and Newton’s methods are ado...
The solution of differential equations is an important problem that arises in a host of areas. Many differential equations are too difficult to solve in closed form. Instead, it becomes necessary to employ numerical techniques. Differential equations have a major application in understanding physical systems that involve aerodynamics, fluid dynamics, thermodynamics, heat diffusion, mechanical o...
exact solution for steady two-dimensional flow of an incompressible magma is obtained. themagmatic flow is studied by considering the magma as a second grade fluid. the governing partialdifferential equations are transformed to ordinary differential equations by symmetry transformations. resultsare discussed through graphs to understand the rheology of the flowing magma
The diierential equation dx(t) = a(x(t); t) dZ (t) + b(x(t); t) dt for fractal-type functions Z (t) is determined via fractional calculus. Under appropriate conditions we prove existence and uniqueness of a local solution by means of its representation x(t) = h(y(t) +Z(t); t) for certain C 1-functions h and y. The method is also applied to It^ o stochastic diierential equations and leads to a g...
which, again, we can solve using a Cauchy integral and (31). The problem of finding K± is more delicate. At first sight, we could take the logarithm of (33), giving logK+− logK− = logG. This looks similar to (31), but it usually happens that logG(t) is not continuous for all t ∈ C , which means that we cannot use (30). However, this difficulty can be overcome. The problem of finding K± such tha...
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