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

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

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

(1 + x^{2} + y^{2} + x^{2} y^{2}) d y = (1 + y^{2}) d x

y = arctan (x) + c_{1}

Solution steps

(1 + x^{2} + y^{2} + x^{2} y^{2}) d y = (1 + y^{2}) d x

Let

y

be the dependent variable. Divide by

d x

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

Substitute

\frac{d y}{d x}

with

y^{'} (x)

(1 + x^{2} + y^{2} + x^{2} y^{2}) y^{^{'}} (x) = 1 + y^{2}

Isolate

y^{'} (x) : y^{'} (x) = \frac{1}{x ^{2} + 1}

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

then

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

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

\int \frac{1}{x ^{2} + 1} d x = arctan (x) + c_{1}

y = arctan (x) + c_{1}

\int \frac{100}{100 - y} d y

y^{^{'''}} + 4 \cdot y^{^{'}} = 3 x - 1

x \to 1 lim (e^{4 x^{5} - 4 x})

\int (2 x - 1)^{2} d x

\frac{\partial}{\partial x} (x^{2} y^{3} + 2 x^{4} y)

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