方程a(t-τ)+b(t)+cx(t-τ)+dx(t)=t的部分解讨论

The Discussion of Some Solutions of The Equation a (t-τ)+b (t)+cx(t-τ)+dx(t)=t

寻求中立型微分方程解的表达式是较困难的,虽有步长法,但在求解中立型微分方程解的表达式方面,至今没有很好的结果.本文用待定系数法讨论了方程:ax·(t-τ)+bx·(t)+cx(t-τ)+dx(t)=t的部分解.

It is difficult to find the expression of solution of neutral differential difference equation.Though there is steplength method,there is not any good result now on the expression of solution of neutral differential difference equation. This paper discusses the expressions of some solution of equation a?x·(t-τ)+b?x·(t)+cx(t-τ)+dx(t)=t with the method of undetermined coefficients.