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Solve log((dy)/(dx))=4x-2y-2 , given tha...

Solve `log((dy)/(dx))=4x-2y-2` , given that `y=1` when `x=1.`

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`log_(e)(dy)/(dx)=4x-2y-2`
`rArr (dy)/(dx)=e^(4x-2y-2)`
`rArr e^(2y+2)dy=e^(4x)dx`
Integrating both sides,
`inte^(2y+2)dy=e^(4x)dx`
`rArr e^(2y_2)/(2)=e^(4x)/(4)+C`
Putting x=1 and y=1, we get
`e^(4)/2=e^(4)/4+C` or `c=e^(4)/4`
So, particular solution is `(e^(2y)+2)/(2) = (e^(4x)+e^(4))/(4)`
or `2(e^(2y+2))=e^(4x)+e^(4)`
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