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

The solution for y^{''}+9y=8x,y(0)=1,y^'(0)=3 is y=cos(3x)+19/27 sin(3x)+(8x)/9

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y^{''}+9y=8x,y(0)=1,y^'(0)=3

y^{^{''}} + 9 y = 8 x, y (0) = 1, y^{^{'}} (0) = 3

y = cos (3 x) + \frac{19}{27} sin (3 x) + \frac{8 x}{9}

Solution steps

y^{''} (x) + 9 y = 8 x

Solve linear ODE:

y = cos (3 x) + \frac{19}{27} sin (3 x) + \frac{8 x}{9}

y = cos (3 x) + \frac{19}{27} sin (3 x) + \frac{8 x}{9}

y^{^{''}} - 8 y^{^{'}} + 16 y = 20 x + 7

y^{^{''}} - y = sin (x) - e^{2 x}

y^{^{''}} + 4 y^{^{'}} - 2 y = 2 t^{2} - 3 t + 6

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} + 9 y = x e^{x}

y^{^{'''}} - y^{^{'}} = csc (t)

What is the solution for y^{''}+9y=8x,y(0)=1,y^'(0)=3 ?
The solution for y^{''}+9y=8x,y(0)=1,y^'(0)=3 is y=cos(3x)+19/27 sin(3x)+(8x)/9