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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

4 y^{^{''}} + 36 y = sec (3 x)

y^{^{''}} - 4 y^{^{'}} - 5 y = 24 x e^{5 x} - 26 sin (x)

(d^{2} + 9) y = 5 e^{x} - 162 x

y^{^{''}} - 3 y^{^{'}} - 9 y = 4 cos (2 x) - 5 e^{- x}

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