What is the solution for y^{''}+16y=6x,y(0)=1,y^'(0)=4 ?

The solution for y^{''}+16y=6x,y(0)=1,y^'(0)=4 is y=cos(4x)+29/32 sin(4x)+(3x)/8

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y^{''}+16y=6x,y(0)=1,y^'(0)=4

y^{^{''}} + 16 y = 6 x, y (0) = 1, y^{^{'}} (0) = 4

y = cos (4 x) + \frac{29}{32} sin (4 x) + \frac{3 x}{8}

Solution steps

y^{''} (x) + 16 y = 6 x

Solve linear ODE:

y = cos (4 x) + \frac{29}{32} sin (4 x) + \frac{3 x}{8}

y = cos (4 x) + \frac{29}{32} sin (4 x) + \frac{3 x}{8}

(D^{3} - 2 D^{2} - 3 D) y = 74 e^{- 3 x} cos (x)

y^{^{''}} - 2 y^{^{'}} - 3 y = - 11 t e^{- t}

y^{^{''}} - 4 y^{^{'}} + 8 y = x^{3} - x

y^{^{''}} - 4 y^{^{'}} + 4 y = \frac{1}{x} e^{2 x}

y^{^{''}} - 4 y^{^{'}} + 4 y = \frac{e ^{- 2 x}}{x ^{2}}

What is the solution for y^{''}+16y=6x,y(0)=1,y^'(0)=4 ?
The solution for y^{''}+16y=6x,y(0)=1,y^'(0)=4 is y=cos(4x)+29/32 sin(4x)+(3x)/8