What is the solution for (3du)/(dt)=u^2 ?

The solution for (3du)/(dt)=u^2 is t=-3/u+c_{1}

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(3du)/(dt)=u^2

\frac{3 d u}{d t} = u^{2}

t = - \frac{3}{u} + c_{1}

Solution steps

\frac{3 d u}{d t} = u^{2}

Let

t

be the dependent variable. Divide by

d u

\frac{3}{\frac{d t}{d u}} = u^{2}

Substitute

\frac{d t}{d u}

with

t^{'} (u)

\frac{3}{t ^{^{'}} ( u )} = u^{2}

Isolate

t^{'} (u) : t^{'} (u) = \frac{3}{u ^{2}}

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

then

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

t = \int \frac{3}{u ^{2}} d u

\int \frac{3}{u ^{2}} d u = - \frac{3}{u} + c_{1}

t = - \frac{3}{u} + c_{1}

t an g e n t y = \frac{x ^{3}}{x + 3}, (3, \frac{9}{2})

\int \frac{2 cos ( x )}{sin ^{3} ( x )} d x

\int 2 x (x^{2} + 5)^{6} d x

\int \frac{1}{( x + y + z + 1 ) ^{3}} d z

x \to 0 lim (\frac{2. 7 ^{x} - 1}{x})

What is the solution for (3du)/(dt)=u^2 ?
The solution for (3du)/(dt)=u^2 is t=-3/u+c_{1}