What is the solution for (D-1)y=2sin(5t),y(0)=0 ?

The solution for (D-1)y=2sin(5t),y(0)=0 is y= 5/13 e^t-1/13 sin(5t)-5/13 cos(5t)

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(D-1)y=2sin(5t),y(0)=0

(D - 1) y = 2 sin (5 t), y (0) = 0

y = \frac{5}{13} e^{t} - \frac{1}{13} sin (5 t) - \frac{5}{13} cos (5 t)

Solution steps

(D - 1) y = 2 sin (5 t)

Solve linear ODE:

y = \frac{5}{13} e^{t} - \frac{1}{13} sin (5 t) - \frac{5}{13} cos (5 t)

y = \frac{5}{13} e^{t} - \frac{1}{13} sin (5 t) - \frac{5}{13} cos (5 t)

x^{^{''}} - 3 x^{^{'}} + x = - e^{t}

y^{^{''}} - 4 y^{^{'}} = 1

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

y^{^{'''}} - 5 y^{^{''}} + 6 y^{^{'}} = 6 e^{x}

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

What is the solution for (D-1)y=2sin(5t),y(0)=0 ?
The solution for (D-1)y=2sin(5t),y(0)=0 is y= 5/13 e^t-1/13 sin(5t)-5/13 cos(5t)