Question

y'=(y+4x)^2

Differential equations
ANSWERED
asked 2021-07-05
\(\displaystyle{y}'={\left({y}+{4}{x}\right)}^{{2}}\)

Answers (1)

2021-07-06
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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