What is the solution for y^{''}+y=3cos(4t),y(0)=1,y^'(0)=2 ?

The solution for y^{''}+y=3cos(4t),y(0)=1,y^'(0)=2 is y= 6/5 cos(t)+2sin(t)-1/5 cos(4t)

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y^{''}+y=3cos(4t),y(0)=1,y^'(0)=2

y^{^{''}} + y = 3 cos (4 t), y (0) = 1, y^{^{'}} (0) = 2

y = \frac{6}{5} cos (t) + 2 sin (t) - \frac{1}{5} cos (4 t)

Solution steps

y^{''} (t) + y = 3 cos (4 t)

Solve linear ODE:

y = \frac{6}{5} cos (t) + 2 sin (t) - \frac{1}{5} cos (4 t)

y = \frac{6}{5} cos (t) + 2 sin (t) - \frac{1}{5} cos (4 t)

y^{^{''}} - 7 y^{^{'}} + 12 y = - e^{4 x}

x^{^{''}} + x = \frac{1}{cos ( t )}

y^{^{''}} - 6 y^{^{'}} + 10 y = e^{3 x} cos (x)

y^{^{''}} + y = x^{2} + sin (x)

y^{^{''}} - 4 y^{^{'}} - 21 y = 320 e^{- t}

What is the solution for y^{''}+y=3cos(4t),y(0)=1,y^'(0)=2 ?
The solution for y^{''}+y=3cos(4t),y(0)=1,y^'(0)=2 is y= 6/5 cos(t)+2sin(t)-1/5 cos(4t)