\frac{\partial ^2u}{\partial \:t^2}=16\frac{\partial ^2u}{\partial \:x^2}\:,\:u\left(x,\:0\right)=e^x,\:\frac{∂u}{∂t}\:\left(x,\:0\right)\:=\:e^{-x}

asked by guest
on Jul 06, 2024 at 6:53 am



Mathbot Says...

I wasn't able to parse your question, but the HE.NET team is hard at work making me smarter.