What is the solution for 2y^'sin(t)-e^tysin(t)=e^tysin(t) ?

The solution for 2y^'sin(t)-e^tysin(t)=e^tysin(t) is y=e^{e^t+c_{1}}

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2y^'sin(t)-e^tysin(t)=e^tysin(t)

2 y^{^{'}} sin (t) - e^{t} y sin (t) = e^{t} y sin (t)

y = e^{e^{t} + c_{1}}

Solution steps

2 y^{'} (t) sin (t) - e^{t} y sin (t) = e^{t} y sin (t)

Solve separable ODE:

y = e^{e^{t} + c_{1}}

y = e^{e^{t} + c_{1}}

\frac{d y}{d t} = \frac{40}{2000} - \frac{y}{200}

y^{^{'}} + 8 y^{3} = 0

\frac{\frac{d y}{d t}}{y - 49} = - \frac{1}{5}

y^{^{'}} (s) (- s^{2} - 2 s) - 2 sy (s) = 0

\frac{d y}{d x} y = (csc (x) - cot (x))^{- 1}

What is the solution for 2y^'sin(t)-e^tysin(t)=e^tysin(t) ?
The solution for 2y^'sin(t)-e^tysin(t)=e^tysin(t) is y=e^{e^t+c_{1}}