What is the solution for (dy)/(dx)=(x+y+3)^2 ?

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

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(dy)/(dx)=(x+y+3)^2

\frac{d y}{d x} = (x + y + 3)^{2}

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

Solution steps

\frac{d y}{d x} = (x + y + 3)^{2}

Solve using substitution:

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

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

\int_{- 3}^{- 1} 2 x^{2} + 3 d x

e^{p} (p + p)

\int \frac{t}{( t + 1 ) ^{4}} d t

x \cdot ln (x) = y (1 + 3 + y^{2}) y^{^{'}} (x), y (1) = 1

\frac{d y}{d x} = 0.0017 y (3000 - y)

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