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

The solution for y^{''}+y=cos(3t),y(0)=1,y^'(0)=0 is y= 9/8 cos(t)-1/8 cos(3t)

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

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

y = \frac{9}{8} cos (t) - \frac{1}{8} cos (3 t)

Solution steps

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

Solve linear ODE:

y = \frac{9}{8} cos (t) - \frac{1}{8} cos (3 t)

y = \frac{9}{8} cos (t) - \frac{1}{8} cos (3 t)

f (x) = 8 x

x^{4} - 13 x^{2} + 36

\frac{d}{d x} (\frac{2 x ^{2}}{6 x + 1})

x \to \infty lim (\frac{x ^{2} + 2 ^{x}}{( x + 1 ) ^{2} + 2 ^{x + 1}})

\frac{d}{d t} (\frac{3 e ^{3 α t} xα}{4})

What is the solution for y^{''}+y=cos(3t),y(0)=1,y^'(0)=0 ?
The solution for y^{''}+y=cos(3t),y(0)=1,y^'(0)=0 is y= 9/8 cos(t)-1/8 cos(3t)