MathDB
Problem 78

Source:

July 1, 2010
trigonometry

Problem Statement

Solve the Cauchy problem
2At2=92Ax22B,  2Bt2=62Bx22A\frac{\partial ^2A}{\partial t^2}=9\frac{\partial^2 A}{\partial x^2}-2B,\;\frac{\partial^2 B}{\partial t^2}=6\frac{\partial^2 B}{\partial x^2}-2A
At=0=cosx,  Bt=0=0,  Att=0=Btt=0=0A|_{t=0}=\cos x,\; B|_{t=0}=0,\; \left.\frac{\partial A}{\partial t}\right|_{t=0}=\left.\frac{\partial B}{\partial t}\right|_{t=0}=0