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

cos (x), 0

\frac{d}{d x} (x^{3} + \frac{4 x ^{2}}{3} - 2 x + 5)

\int x^{3} x^{2} + 36 d x

x \to 0 lim (\frac{( 2 x ^{3} )}{2 x})

\int e^{2 x} \cdot 2 d x

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