What is the solution for y^'=(x+1)/y ?

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

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y^'=(x+1)/y

y^{^{'}} = \frac{x + 1}{y}

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

Solution steps

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

Solve separable ODE:

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

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

x \to \infty lim (3 x + \frac{1}{x})

\int \frac{2 x - 1}{x ^{3} - 6 x ^{2} + 11 x - 6} d x

x \to - \infty lim (- \frac{6}{x})

f (x) = csc^{7} (2 x)

y^{^{'}} + 2 y = e^{(4 t)}

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