What is the solution for (dv}{dt}+\frac{(v^'*v))/t =v^'*(t/2-1/2) ?

The solution for (dv}{dt}+\frac{(v^'*v))/t =v^'*(t/2-1/2) is v=c_{1}

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(dv}{dt}+\frac{(v^'v))/t =v^'(t/2-1/2)

\frac{d v}{d t} + \frac{( v ^{^{'}} \cdot v )}{t} = v^{^{'}} \cdot (\frac{t}{2} - \frac{1}{2})

v = c_{1}

Solution steps

v^{'} (t) + \frac{( v ^{'} ( t ) \cdot v )}{t} = v^{'} (t) \cdot (\frac{t}{2} - \frac{1}{2})

Isolate

v^{'} (t) : v^{'} (t) = 0

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

v = \int 0 d t

\int 0 d t = c_{1}

v = c_{1}

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What is the solution for (dv}{dt}+\frac{(v^'*v))/t =v^'*(t/2-1/2) ?
The solution for (dv}{dt}+\frac{(v^'*v))/t =v^'*(t/2-1/2) is v=c_{1}