What is the solution for y^{''}+9y=sin(2.9t),y(0)=1,y^'(0)=1 ?

The solution for y^{''}+9y=sin(2.9t),y(0)=1,y^'(0)=1 is y=cos(3t)-1.30508…sin(3t)+1.69491…sin(2.9t)

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y^{''}+9y=sin(2.9t),y(0)=1,y^'(0)=1

y^{^{''}} + 9 y = sin (2.9 t), y (0) = 1, y^{^{'}} (0) = 1

y = cos (3 t) - 1.30508 \dots sin (3 t) + 1.69491 \dots sin (2.9 t)

Solution steps

y^{''} (t) + 9 y = sin (2.9 t)

Solve linear ODE:

y = cos (3 t) - 1.30508 \dots sin (3 t) + 1.69491 \dots sin (2.9 t)

y = cos (3 t) - 1.30508 \dots sin (3 t) + 1.69491 \dots sin (2.9 t)

What is the solution for y^{''}+9y=sin(2.9t),y(0)=1,y^'(0)=1 ?
The solution for y^{''}+9y=sin(2.9t),y(0)=1,y^'(0)=1 is y=cos(3t)-1.30508…sin(3t)+1.69491…sin(2.9t)