What is the solution for y^{''}-4y^'+4y=sin(2t),y(0)=1,y^'(0)=1 ?

The solution for y^{''}-4y^'+4y=sin(2t),y(0)=1,y^'(0)=1 is y= 7/8 e^{2t}-(3e^{2t}t)/4+1/8 cos(2t)

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y^{''}-4y^'+4y=sin(2t),y(0)=1,y^'(0)=1

y^{^{''}} - 4 y^{^{'}} + 4 y = sin (2 t), y (0) = 1, y^{^{'}} (0) = 1

y = \frac{7}{8} e^{2 t} - \frac{3 e ^{2 t} t}{4} + \frac{1}{8} cos (2 t)

Solution steps

y^{''} (t) - 4 y^{'} (t) + 4 y = sin (2 t)

Solve linear ODE:

y = \frac{7}{8} e^{2 t} - \frac{3 e ^{2 t} t}{4} + \frac{1}{8} cos (2 t)

y = \frac{7}{8} e^{2 t} - \frac{3 e ^{2 t} t}{4} + \frac{1}{8} cos (2 t)

What is the solution for y^{''}-4y^'+4y=sin(2t),y(0)=1,y^'(0)=1 ?
The solution for y^{''}-4y^'+4y=sin(2t),y(0)=1,y^'(0)=1 is y= 7/8 e^{2t}-(3e^{2t}t)/4+1/8 cos(2t)