y'=(y+4x)^2

texelaare 2021-07-05 Answered
\(\displaystyle{y}'={\left({y}+{4}{x}\right)}^{{2}}\)

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Expert Answer

ensojadasH
Answered 2021-07-06 Author has 6665 answers
The solution of the given Differential equations is \(\displaystyle{y}={2}{\tan{{\left({2}{x}+{c}\right)}}}-{4}{x}\), where c is a constant.
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