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Popular >

(dv)/(dt)=v^2+5v+2

Solution

\frac{d v}{d t} = v^{2} + 5 v + 2

Solution

v = \frac{5 e ^{17 c_{1}}}{2 ( e ^{17 t} - e ^{17 c_{1}} )} + \frac{17 e ^{17 c_{1}}}{2 ( e ^{17 t} - e ^{17 c_{1}} )} - \frac{5 e ^{17 t}}{2 ( e ^{17 t} - e ^{17 c_{1}} )} + \frac{17 e ^{17 t}}{2 ( e ^{17 t} - e ^{17 c_{1}} )}

Solution steps

\frac{d v}{d t} = v^{2} + 5 v + 2

Substitute

\frac{d v}{d t}

with

v^{'} (t)

v^{^{'}} (t) = v^{2} + 5 v + 2

Solve for

t, v^{'} (t) = \frac{d v}{d t} = \frac{1}{\frac{d t}{d v}} = \frac{1}{t ^{^{'}}}

(\frac{1}{t ^{^{^{'}}}}) = v^{2} + 5 v + 2

Isolate

t^{^{'}} : t^{^{'}} = \frac{1}{v ^{2} + 5 v + 2}

If

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

then

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

t = \int \frac{1}{v ^{2} + 5 v + 2} d v

\int \frac{1}{v ^{2} + 5 v + 2} d v = - \frac{1}{17} (ln (\frac{2 v + 5}{17} + 1) - ln (\frac{2 v + 5}{17} - 1)) + c_{1}

t = - \frac{1}{17} (ln (\frac{2 v + 5}{17} + 1) - ln (\frac{2 v + 5}{17} - 1)) + c_{1}

Isolate

v : v = \frac{5 e ^{17 c_{1}}}{2 ( e ^{17 t} - e ^{17 c_{1}} )} + \frac{17 e ^{17 c_{1}}}{2 ( e ^{17 t} - e ^{17 c_{1}} )} - \frac{5 e ^{17 t}}{2 ( e ^{17 t} - e ^{17 c_{1}} )} + \frac{17 e ^{17 t}}{2 ( e ^{17 t} - e ^{17 c_{1}} )}

v = \frac{5 e ^{17 c_{1}}}{2 ( e ^{17 t} - e ^{17 c_{1}} )} + \frac{17 e ^{17 c_{1}}}{2 ( e ^{17 t} - e ^{17 c_{1}} )} - \frac{5 e ^{17 t}}{2 ( e ^{17 t} - e ^{17 c_{1}} )} + \frac{17 e ^{17 t}}{2 ( e ^{17 t} - e ^{17 c_{1}} )}

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Frequently Asked Questions (FAQ)

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