What is the solution for y^'=2y+e^{2x}cos(3x) ?

The solution for y^'=2y+e^{2x}cos(3x) is y=(sin(3x))/(3e^{-2x)}+c_{1}e^{2x}

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y^'=2y+e^{2x}cos(3x)

y^{^{'}} = 2 y + e^{2 x} cos (3 x)

y = \frac{sin ( 3 x )}{3 e ^{- 2 x}} + c_{1} e^{2 x}

Solution steps

y^{'} (x) = 2 y + e^{2 x} cos (3 x)

Solve linear ODE:

y = \frac{sin ( 3 x )}{3 e ^{- 2 x}} + c_{1} e^{2 x}

y = \frac{sin ( 3 x )}{3 e ^{- 2 x}} + c_{1} e^{2 x}

\int y \cdot 25 \cdot (y - Θ) \cdot e^{(- 5 (y - Θ))} d y

\int y^{2} e^{y} d y

\frac{d}{d x} (ln (sin (6 x)))

(- 6, 6), (4, - 7)

3 y^{^{''}} = 2 y^{^{'}}

What is the solution for y^'=2y+e^{2x}cos(3x) ?
The solution for y^'=2y+e^{2x}cos(3x) is y=(sin(3x))/(3e^{-2x)}+c_{1}e^{2x}