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

The solution for 2y(x+1)dy=xdx is y=sqrt(x+1-ln(x+1)+c_{1)},y=-sqrt(x+1-ln(x+1)+c_{1)}

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2y(x+1)dy=xdx

2 y (x + 1) d y = x d x

y = x + 1 - ln (x + 1) + c_{1}, y = - x + 1 - ln (x + 1) + c_{1}

Solution steps

2 y (x + 1) d y = x d x

Solve separable ODE:

y = x + 1 - ln (x + 1) + c_{1}, y = - x + 1 - ln (x + 1) + c_{1}

y = x + 1 - ln (x + 1) + c_{1}, y = - x + 1 - ln (x + 1) + c_{1}

\frac{1}{x ^{2}}, at 4

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

\int_{2}^{3} 3 - x d x

\frac{d}{d x} (ln (x) csc (x))

y^{^{'}} = - 5 sin (3 t) + 2 cos (4 t)

What is the solution for 2y(x+1)dy=xdx ?
The solution for 2y(x+1)dy=xdx is y=sqrt(x+1-ln(x+1)+c_{1)},y=-sqrt(x+1-ln(x+1)+c_{1)}