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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int_{1}^{2} \frac{3}{4 x - x ^{2}} d x

f (x) = x - 4, at x = 13

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

x \to 1 lim (∣ e^{x} - 1 ∣)

\int \frac{2}{3 x} d x

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