What is the solution for (dy)/(dx)=ysin(3x)-3ye^{-3x},y(0)=1 ?

The solution for (dy)/(dx)=ysin(3x)-3ye^{-3x},y(0)=1 is y=e^{-1/3 cos(3x)+e^{-3x}-2/3}

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(dy)/(dx)=ysin(3x)-3ye^{-3x},y(0)=1

\frac{d y}{d x} = y sin (3 x) - 3 y e^{- 3 x}, y (0) = 1

y = e^{- \frac{1}{3} c o s (3 x) + e^{- 3 x} - \frac{2}{3}}

Solution steps

\frac{d y}{d x} = y sin (3 x) - 3 y e^{- 3 x}

Solve separable ODE:

y = e^{- \frac{1}{3} c o s (3 x) + e^{- 3 x} - \frac{2}{3}}

y = e^{- \frac{1}{3} c o s (3 x) + e^{- 3 x} - \frac{2}{3}}

y^{^{'}} + y tan (x) = 0, y (0) = a cos (x)

x^{2} (x d y + y d x) = 8 x d y - 8 y d x

3 y^{^{'}} = - 9 y

y^{^{'}} \frac{2}{x} y = 4 x

y^{2} y^{^{'}} - x^{2} = 0, y (- 2) = - 2

What is the solution for (dy)/(dx)=ysin(3x)-3ye^{-3x},y(0)=1 ?
The solution for (dy)/(dx)=ysin(3x)-3ye^{-3x},y(0)=1 is y=e^{-1/3 cos(3x)+e^{-3x}-2/3}