Q44·Section B (Part B)·44 of 120d'Alembert solution of the 1-D wave equationLet u=u(x,t)u=u(x,t)u=u(x,t) be the solution of ∂2u∂t2−∂2u∂x2=0,x∈R,t>0,u(x,0)=1+x2,∂u∂t(x,0)=x+1.\frac{\partial^2u}{\partial t^2}-\frac{\partial^2u}{\partial x^2}=0, x\in\mathbb R, t>0,\qquad u(x,0)=1+x^2,\qquad \frac{\partial u}{\partial t}(x,0)=x+1.∂t2∂2u−∂x2∂2u=0,x∈R,t>0,u(x,0)=1+x2,∂t∂u(x,0)=x+1. Then the value of u(1,1)u(1,1)u(1,1) is OptionsA2B3C4D5Pick an option to check your answer.Show answerShow solution Report a problem with this question← Q43All questionsQ45 →Keyboard: A–D answer · S solution · ←/→ previous/next