What is the solution for v^'=((v+4/v))/t ?

The solution for v^'=((v+4/v))/t is v(t)=sqrt(c_{1)t^2-4},v(t)=-sqrt(c_{1)t^2-4}

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v^'=((v+4/v))/t

v^{^{'}} = \frac{( v + \frac{4}{v} )}{t}

v (t) = c_{1} t^{2} - 4, v (t) = - c_{1} t^{2} - 4

Solution steps

v^{'} (t) = \frac{( v ( t ) + \frac{4}{v ( t )} )}{t}

Solve separable ODE:

v (t) = c_{1} t^{2} - 4, v (t) = - c_{1} t^{2} - 4

v (t) = c_{1} t^{2} - 4, v (t) = - c_{1} t^{2} - 4

\frac{d y}{d x} 6 x y + 1 = 0

(x^{2} + 6)^{\frac{1}{2}} d x + \frac{1}{x csc ( 8 y )} d y = 0

\frac{d y}{d x} = 72 \cdot y \cdot x^{7}

(x^{2} + 1) \frac{d y}{d x} + x y = 3 x, y (0) = 4

y \frac{d y}{d x} = 6 (y^{2} + 7)^{7}

What is the solution for v^'=((v+4/v))/t ?
The solution for v^'=((v+4/v))/t is v(t)=sqrt(c_{1)t^2-4},v(t)=-sqrt(c_{1)t^2-4}