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

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

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

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

y = \frac{ln ( x ^{2} + 1 ) + c _{1}}{2}, y = - \frac{ln ( x ^{2} + 1 ) + c _{1}}{2}

Solution steps

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

Solve separable ODE:

y = \frac{ln ( x ^{2} + 1 ) + c _{1}}{2}, y = - \frac{ln ( x ^{2} + 1 ) + c _{1}}{2}

y = \frac{ln ( x ^{2} + 1 ) + c _{1}}{2}, y = - \frac{ln ( x ^{2} + 1 ) + c _{1}}{2}

y^{2} e^{x} + 3 y e^{2 x} \frac{d y}{d x} = 2 e^{x}, y (0) = 1

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

y^{^{'}} = 9 y

(x + 1) \frac{d y}{d x} + y = 10

x^{^{'}} + x = - 1

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