What is the solution for y^'=(5x+8)/y ?

The solution for y^'=(5x+8)/y is y=sqrt(5x^2+16x+c_{1)},y=-sqrt(5x^2+16x+c_{1)}

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y^'=(5x+8)/y

y^{^{'}} = \frac{5 x + 8}{y}

y = 5 x^{2} + 16 x + c_{1}, y = - 5 x^{2} + 16 x + c_{1}

Solution steps

y^{'} (x) = \frac{5 x + 8}{y}

Solve separable ODE:

y = 5 x^{2} + 16 x + c_{1}, y = - 5 x^{2} + 16 x + c_{1}

y = 5 x^{2} + 16 x + c_{1}, y = - 5 x^{2} + 16 x + c_{1}

\frac{d y}{d x} = 2 x 36 - y^{2}

\frac{d y}{d x} = \frac{1}{7} (y)^{\frac{1}{2}} (cos (y^{\frac{1}{2}}))^{2}

y^{^{'}} = \frac{y ^{4}}{x ^{6}}, y (1) = 3

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

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

What is the solution for y^'=(5x+8)/y ?
The solution for y^'=(5x+8)/y is y=sqrt(5x^2+16x+c_{1)},y=-sqrt(5x^2+16x+c_{1)}