What is the solution for y^{''}+9y=3x,y(0)=1,y^'(0)=3 ?

The solution for y^{''}+9y=3x,y(0)=1,y^'(0)=3 is y=cos(3x)+8/9 sin(3x)+x/3

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y^{''}+9y=3x,y(0)=1,y^'(0)=3

y^{^{''}} + 9 y = 3 x, y (0) = 1, y^{^{'}} (0) = 3

y = cos (3 x) + \frac{8}{9} sin (3 x) + \frac{x}{3}

Solution steps

y^{''} (x) + 9 y = 3 x

Solve linear ODE:

y = cos (3 x) + \frac{8}{9} sin (3 x) + \frac{x}{3}

y = cos (3 x) + \frac{8}{9} sin (3 x) + \frac{x}{3}

y^{^{'''}} + 8 y^{^{''}} - 13 y^{^{'}} - 140 y = 4 e^{4} t

y^{^{''''}} - y = e^{x}

y^{^{''}} - 35 y^{^{'}} + 9 y = 60 t, y (0) = - 1, y^{^{'}} (0) = 2

y^{^{''}} + 2 y^{^{'}} + 5 y = 52 cos (3 x)

y^{^{''}} + 3 y^{^{'}} + 5 = 0

What is the solution for y^{''}+9y=3x,y(0)=1,y^'(0)=3 ?
The solution for y^{''}+9y=3x,y(0)=1,y^'(0)=3 is y=cos(3x)+8/9 sin(3x)+x/3