"y''+16y'=8\\cos4t\\\\\ny(0)=0, y'(0)=8\\\\"
Laplace transforms is
"y(t)\\mapsto Y\\\\\ny''(t)\\mapsto p^2Y-py(0)-y'(0)=\\\\\n=p^2Y-p\u20220-8=p^2Y-8 \\\\\n\\cos4t\\mapsto\\frac{p}{p^2+16}"
then
"p^2Y-8+16Y=\\frac{8p}{p^2+16}\\\\\n(p^2+16)Y=\\frac{8p}{p^2+16}+8\\\\\nY=\\frac{8p}{(p^2+16)^2}+\\frac{8}{p^2+16}"
Then
"\\frac{8p}{(p^2+16)^2}=\\\\\n=\\frac{2p\u20224}{(p^2+4^2)^2}\\mapsto t\\sin 4t\\\\"
"\\frac{8}{p^2+16}=\\frac{2\u20224}{p^2+4^2}\\mapsto 2\\sin 4t"
Solution is
"y=t\\sin4t+2\\sin4t\\\\\ny=(t+2)\\sin4t"
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