\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}
Mathbot Says...
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