What is the solution for y^{''}+4y=e^{3x} ?

The solution for y^{''}+4y=e^{3x} is y=c_{1}cos(2x)+c_{2}sin(2x)+1/13 e^{3x}

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y^{''}+4y=e^{3x}

y^{^{''}} + 4 y = e^{3 x}

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{1}{13} e^{3 x}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{1}{13} e^{3 x}

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{1}{13} e^{3 x}

y^{^{''}} + 4 y^{^{'}} + 4 y = 2 x + 5

y^{^{'''}} - 2 y^{^{'}} + 4 y = e^{x} cos (x) + x^{2} + sin (2 x)

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

y^{^{''}} + b^{2} y = b

y^{^{'''}} + 225 y^{^{'}} = tan (15 x)

What is the solution for y^{''}+4y=e^{3x} ?
The solution for y^{''}+4y=e^{3x} is y=c_{1}cos(2x)+c_{2}sin(2x)+1/13 e^{3x}