What is the solution for (dy)/(dx)+y^2=1 ?

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

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(dy)/(dx)+y^2=1

\frac{d y}{d x} + y^{2} = 1

y = - \frac{e ^{2 x + 2 c_{1}}}{1 - e ^{2 (x + c_{1})}} - \frac{1}{1 - e ^{2 (x + c_{1})}}

Solution steps

\frac{d y}{d x} + y^{2} = 1

Solve separable ODE:

y = - \frac{e ^{2 x + 2 c_{1}}}{1 - e ^{2 (x + c_{1})}} - \frac{1}{1 - e ^{2 (x + c_{1})}}

y = - \frac{e ^{2 x + 2 c_{1}}}{1 - e ^{2 (x + c_{1})}} - \frac{1}{1 - e ^{2 (x + c_{1})}}

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

\int e^{s i n (θ)} cos (θ) d θ

f (x) = \frac{5 x}{4 x ^{2} + 1}

\int \frac{x + 4}{x ^{2} + 1} d x

f (t) = t \cdot sin (3 t)

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