What is the solution for derivative of (sin(pix+cos(pi)y)^2)=2 ?

The solution for derivative of (sin(pix+cos(pi)y)^2)=2 is y=-sqrt(2x+c_{1)},y=sqrt(2x+c_{1)}

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derivative of (sin(pix+cos(pi)y)^2)=2

\frac{d}{d x} ((sin (π) x + cos (π) y)^{2}) = 2

y = - 2 x + c_{1}, y = 2 x + c_{1}

Solution steps

\frac{d}{d x} ((sin (π) x + cos (π) y)^{2}) = 2

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

(sin (π) x + cos (π) y)^{2} = \int 2 d x

\int 2 d x = 2 x + c_{1}

(sin (π) x + cos (π) y)^{2} = 2 x + c_{1}

Isolate

y : y = - 2 x + c_{1}, y = 2 x + c_{1}

y = - 2 x + c_{1}, y = 2 x + c_{1}

y^{^{'}} = e^{- y} - 1

y^{^{'}} + y = 1, - H (t - 2) y (0) = 2

7 x y y^{^{'}} = x + 6 x^{2} - 8

y^{^{'}} = 20 y - 25

\frac{d y}{d x} + k y = 0

What is the solution for derivative of (sin(pix+cos(pi)y)^2)=2 ?
The solution for derivative of (sin(pix+cos(pi)y)^2)=2 is y=-sqrt(2x+c_{1)},y=sqrt(2x+c_{1)}