What is the solution for y^{''}-2y^'+5y=4e^{3t},y(0)=0,y^'(0)=-3 ?

The solution for y^{''}-2y^'+5y=4e^{3t},y(0)=0,y^'(0)=-3 is y=e^t(-1/2 cos(2t)-2sin(2t))+1/2 e^{3t}

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y^{''}-2y^'+5y=4e^{3t},y(0)=0,y^'(0)=-3

y^{^{''}} - 2 y^{^{'}} + 5 y = 4 e^{3 t}, y (0) = 0, y^{^{'}} (0) = - 3

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

Solution steps

y^{''} (t) - 2 y^{'} (t) + 5 y = 4 e^{3 t}

Solve linear ODE:

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

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

What is the solution for y^{''}-2y^'+5y=4e^{3t},y(0)=0,y^'(0)=-3 ?
The solution for y^{''}-2y^'+5y=4e^{3t},y(0)=0,y^'(0)=-3 is y=e^t(-1/2 cos(2t)-2sin(2t))+1/2 e^{3t}