What is the solution for y^{''}-6y^'+13y=4e^{3x},y(0)=2,y^'(0)=4 ?

The solution for y^{''}-6y^'+13y=4e^{3x},y(0)=2,y^'(0)=4 is y=e^{3x}(cos(2x)-sin(2x))+e^{3x}

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y^{''}-6y^'+13y=4e^{3x},y(0)=2,y^'(0)=4

y^{^{''}} - 6 y^{^{'}} + 13 y = 4 e^{3 x}, y (0) = 2, y^{^{'}} (0) = 4

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

Solution steps

y^{''} (x) - 6 y^{'} (x) + 13 y = 4 e^{3 x}

Solve linear ODE:

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

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

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

y^{^{''}} - y = e^{t}, y (0) = 1, y^{^{'}} (0) = 1

y^{^{''}} - 4 y^{^{'}} + 13 y = 3 e^{2 x} - 5 e^{3 x}

y^{^{''}} - y = cosh (3 x)

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

What is the solution for y^{''}-6y^'+13y=4e^{3x},y(0)=2,y^'(0)=4 ?
The solution for y^{''}-6y^'+13y=4e^{3x},y(0)=2,y^'(0)=4 is y=e^{3x}(cos(2x)-sin(2x))+e^{3x}