What is the solution for (dx)/(dt)=sin(3t)cos(3t) ?

The solution for (dx)/(dt)=sin(3t)cos(3t) is x=-1/6 cos^2(3t)+c_{1}

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(dx)/(dt)=sin(3t)cos(3t)

\frac{d x}{d t} = sin (3 t) cos (3 t)

x = - \frac{1}{6} cos^{2} (3 t) + c_{1}

Solution steps

\frac{d x}{d t} = sin (3 t) cos (3 t)

Substitute

\frac{d x}{d t}

with

x^{'} (t)

x^{^{'}} (t) = sin (3 t) cos (3 t)

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

then

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

x = \int sin (3 t) cos (3 t) d t

\int sin (3 t) cos (3 t) d t = - \frac{1}{6} cos^{2} (3 t) + c_{1}

x = - \frac{1}{6} cos^{2} (3 t) + c_{1}

x^{3} - 17 x^{2} + 63 x + 6

\frac{\partial}{\partial x} (x^{2} + y^{2} + 3 z)

\int (2 x) cos (x) d x

tan (x), 0, 1, 2, 3, \frac{π}{4}

What is the solution for (dx)/(dt)=sin(3t)cos(3t) ?
The solution for (dx)/(dt)=sin(3t)cos(3t) is x=-1/6 cos^2(3t)+c_{1}