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

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

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y^'=(2x+y)^2

y^{^{'}} = (2 x + y)^{2}

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

Solution steps

y^{'} (x) = (2 x + y)^{2}

Solve using substitution:

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

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

y^{^{''}} + 7 y + 10 = 0, y (0) = 3, y^{^{'}} (0) = 18

\frac{d}{d s} (\frac{6 s ^{3} + 28 s ^{2} + 38 s + 15}{( s + 1 ) ^{2}})

\frac{d}{d x} (2 x^{3} - 9 x^{2} + 12 x - 3)

f (x) = \frac{1}{x ^{8}}

\frac{d}{d x} (ln (ln (15 x)))

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