What is the solution for linear y^{''}-2y^'+5y=6e^t,y(0)=0 ?

The solution for linear y^{''}-2y^'+5y=6e^t,y(0)=0 is y=e^t(-3/2 cos(2t)+c_{2}sin(2t))+3/2 e^t

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linear y^{''}-2y^'+5y=6e^t,y(0)=0

linear

y^{^{''}} - 2 y^{^{'}} + 5 y = 6 e^{t}, y (0) = 0

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

Solution steps

y^{''} (t) - 2 y^{'} (t) + 5 y = 6 e^{t}

Solve linear ODE:

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

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

y^{^{''}} + 3 y^{^{'}} + 2 y = 10 u, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 2 y^{^{'}} - 2 = 0

\frac{d ^{2} y}{d x ^{2}} - y - 6 = 1

y^{^{''}} - 2 y^{^{'}} + 2 y = 2 e^{t}

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 9 y = x e^{x}

What is the solution for linear y^{''}-2y^'+5y=6e^t,y(0)=0 ?
The solution for linear y^{''}-2y^'+5y=6e^t,y(0)=0 is y=e^t(-3/2 cos(2t)+c_{2}sin(2t))+3/2 e^t