What is the solution for v^'=sqrt(2v),v(0)=3.2 ?

The solution for v^'=sqrt(2v),v(0)=3.2 is v(t)=0.5t^2+2.52982…t+3.2

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v^'=sqrt(2v),v(0)=3.2

v^{^{'}} = 2 v, v (0) = 3.2

v (t) = 0.5 t^{2} + 2.52982 \dots t + 3.2

Solution steps

v^{'} (t) = 2 v (t)

Solve separable ODE:

v (t) = 0.5 t^{2} + 2.52982 \dots t + 3.2

v (t) = 0.5 t^{2} + 2.52982 \dots t + 3.2

\frac{d y}{d x} - \frac{y}{4 x} = 0

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

\frac{d x}{d t} = e^{x} \cdot t \cdot cos (t)

y^{^{'}} = \frac{4 t y}{2 + t ^{2}}, y (0) = 4

e^{y} y^{^{'}} - x^{2} = 0, y (0) = 0

What is the solution for v^'=sqrt(2v),v(0)=3.2 ?
The solution for v^'=sqrt(2v),v(0)=3.2 is v(t)=0.5t^2+2.52982…t+3.2