What is the solution for dx+sin(t)xdtsin(t)dt=0 ?

The solution for dx+sin(t)xdtsin(t)dt=0 is x=e^{-1/2 (t-1/2 sin(2t))+c_{1}}

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dx+sin(t)xdtsin(t)dt=0

d x + sin (t) x d t sin (t) d t = 0

x = e^{- \frac{1}{2} (t - \frac{1}{2} s i n (2 t)) + c_{1}}

Solution steps

d x + sin (t) x d t sin (t) d t = 0

Solve separable ODE:

x = e^{- \frac{1}{2} (t - \frac{1}{2} s i n (2 t)) + c_{1}}

x = e^{- \frac{1}{2} (t - \frac{1}{2} s i n (2 t)) + c_{1}}

y^{^{'}} = \frac{5 + 12 x}{x y ^{2}}, y (1) = 3

\frac{d y}{d x} = y (\frac{1}{x}) + y

y^{^{'}} = \frac{1 + 2 y}{4 - x ^{2}}, y (1) = 0

\frac{d v}{d t} = 5 - 4 v

\frac{d y}{d x} y = sinh (11 x)

What is the solution for dx+sin(t)xdtsin(t)dt=0 ?
The solution for dx+sin(t)xdtsin(t)dt=0 is x=e^{-1/2 (t-1/2 sin(2t))+c_{1}}