What is the solution for y^'=2(3x+y)^2-1 ?

The solution for y^'=2(3x+y)^2-1 is y=tan(2x+c_{1})-3x

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y^'=2(3x+y)^2-1

y^{^{'}} = 2 (3 x + y)^{2} - 1

y = tan (2 x + c_{1}) - 3 x

Solution steps

y^{'} (x) = 2 (3 x + y)^{2} - 1

Solve using substitution:

y = tan (2 x + c_{1}) - 3 x

y = tan (2 x + c_{1}) - 3 x

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What is the solution for y^'=2(3x+y)^2-1 ?
The solution for y^'=2(3x+y)^2-1 is y=tan(2x+c_{1})-3x