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

The solution for y^{''}-3y^'=2e^xsin(x) is y=c_{1}e^{3x}+c_{2}-3/5 e^xsin(x)+1/5 e^xcos(x)

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y^{''}-3y^'=2e^xsin(x)

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

y = c_{1} e^{3 x} + c_{2} - \frac{3}{5} e^{x} sin (x) + \frac{1}{5} e^{x} cos (x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} - \frac{3}{5} e^{x} sin (x) + \frac{1}{5} e^{x} cos (x)

y = c_{1} e^{3 x} + c_{2} - \frac{3}{5} e^{x} sin (x) + \frac{1}{5} e^{x} cos (x)

y^{^{''}} - y^{^{'}} + y = sin (t)

y^{^{''}} + 2 y^{^{'}} + 3 y = 12 e^{x} + 3 x

y^{^{''}} - y = (1 - e^{2 x})^{- \frac{3}{2}}

y^{^{''}} + 4 y^{^{'}} - 2 y = x^{3}

(d^{2} + d - 6) y = e^{- 3} x

What is the solution for y^{''}-3y^'=2e^xsin(x) ?
The solution for y^{''}-3y^'=2e^xsin(x) is y=c_{1}e^{3x}+c_{2}-3/5 e^xsin(x)+1/5 e^xcos(x)