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

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

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y^'=(pix+y+pi^2)^2

y^{^{'}} = (π x + y + π^{2})^{2}

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

Solution steps

y^{'} (x) = (π x + y + π^{2})^{2}

Solve using substitution:

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

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

\frac{1}{x - 1}, 13

x \to 0 lim (x^{5})

\frac{\partial}{\partial x} (12 + 5 x y)

\int sin^{2} (π x) d x

sin (- 2 t)

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