What is the solution for d/(dt)(v(t)e^{kt})=ge^{kt} ?

The solution for d/(dt)(v(t)e^{kt})=ge^{kt} is v(t)= g/k+(c_{1})/(e^{kt)}

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d/(dt)(v(t)e^{kt})=ge^{kt}

\frac{d}{d t} (v (t) e^{k t}) = g e^{k t}

v (t) = \frac{g}{k} + \frac{c _{1}}{e ^{k t}}

Solution steps

\frac{d}{d t} (v (t) e^{k t}) = g e^{k t}

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

then

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

v (t) e^{k t} = \int g e^{k t} d t

\int g e^{k t} d t = g \frac{1}{k} e^{k t} + c_{1}

v (t) e^{k t} = g \frac{1}{k} e^{k t} + c_{1}

Simplify

v (t) e^{k t} = \frac{g e ^{k t}}{k} + c_{1}

Isolate

v (t) : v (t) = \frac{g}{k} + \frac{c _{1}}{e ^{k t}}

v (t) = \frac{g}{k} + \frac{c _{1}}{e ^{k t}}

d x = x^{7} y d y

\frac{d y}{d x} = (\frac{( ln ( x ^{6} ) )}{x y ^{2}}), y (1) = 5

\frac{d y}{d x} 5 x^{2} - 4 y^{2} = 16

\frac{d y}{d x} y = e^{x + y}

(x y + y^{2}) y^{^{'}} = x^{2} + x y, y (1) = 0

What is the solution for d/(dt)(v(t)e^{kt})=ge^{kt} ?
The solution for d/(dt)(v(t)e^{kt})=ge^{kt} is v(t)= g/k+(c_{1})/(e^{kt)}