What is the solution for y^{''}+4y=cos(x),y(0)=6,y^'(0)=6 ?

The solution for y^{''}+4y=cos(x),y(0)=6,y^'(0)=6 is y= 17/3 cos(2x)+3sin(2x)+1/3 cos(x)

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y^{''}+4y=cos(x),y(0)=6,y^'(0)=6

y^{^{''}} + 4 y = cos (x), y (0) = 6, y^{^{'}} (0) = 6

y = \frac{17}{3} cos (2 x) + 3 sin (2 x) + \frac{1}{3} cos (x)

Solution steps

y^{''} (x) + 4 y = cos (x)

Solve linear ODE:

y = \frac{17}{3} cos (2 x) + 3 sin (2 x) + \frac{1}{3} cos (x)

y = \frac{17}{3} cos (2 x) + 3 sin (2 x) + \frac{1}{3} cos (x)

y^{^{''}} + 6 y^{^{'}} + 8 y = 3 t e^{- 4 t}

3 x^{^{''}} + 48 x = 120 e^{- 4 t} (cos (4 t))

D, D (t) = \frac{100 ln ( t + 2 )}{t - 2}

\frac{d ^{2} y}{d t ^{2}} - 2 \frac{d y}{d t} - 24 y = 4 e^{6 t}

y^{^{''}} - y^{^{'}} = cos (x) + 4 e^{- x}

What is the solution for y^{''}+4y=cos(x),y(0)=6,y^'(0)=6 ?
The solution for y^{''}+4y=cos(x),y(0)=6,y^'(0)=6 is y= 17/3 cos(2x)+3sin(2x)+1/3 cos(x)