What is the solution for (xy+2y+x+2)dy-(x^2-4)dx=0 ?

The solution for (xy+2y+x+2)dy-(x^2-4)dx=0 is y=-1+sqrt(x^2-4x+c_{1)+1},y=-sqrt(x^2-4x+c_{1)+1}-1

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(xy+2y+x+2)dy-(x^2-4)dx=0

(x y + 2 y + x + 2) d y - (x^{2} - 4) d x = 0

y = - 1 + x^{2} - 4 x + c_{1} + 1, y = - x^{2} - 4 x + c_{1} + 1 - 1

Solution steps

(x y + 2 y + x + 2) d y - (x^{2} - 4) d x = 0

Solve separable ODE:

y = - 1 + x^{2} - 4 x + c_{1} + 1, y = - x^{2} - 4 x + c_{1} + 1 - 1

y = - 1 + x^{2} - 4 x + c_{1} + 1, y = - x^{2} - 4 x + c_{1} + 1 - 1

y^{^{'}} = \frac{3 x ^{2} + 4 x + 2}{2 y + 1}

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

y^{^{'}} = e^{t + y}, y (ln (2)) = 1

4 y d y = \frac{3}{x} d x

\frac{d y}{d x} x^{2} y - y^{2} = 8

What is the solution for (xy+2y+x+2)dy-(x^2-4)dx=0 ?
The solution for (xy+2y+x+2)dy-(x^2-4)dx=0 is y=-1+sqrt(x^2-4x+c_{1)+1},y=-sqrt(x^2-4x+c_{1)+1}-1