What is the solution for y^{''}+9y=t+3,y(0)=1,y^'(0)=2 ?

The solution for y^{''}+9y=t+3,y(0)=1,y^'(0)=2 is y= 2/3 cos(3t)+17/27 sin(3t)+t/9+1/3

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y^{''}+9y=t+3,y(0)=1,y^'(0)=2

y^{^{''}} + 9 y = t + 3, y (0) = 1, y^{^{'}} (0) = 2

y = \frac{2}{3} cos (3 t) + \frac{17}{27} sin (3 t) + \frac{t}{9} + \frac{1}{3}

Solution steps

y^{''} (t) + 9 y = t + 3

Solve linear ODE:

y = \frac{2}{3} cos (3 t) + \frac{17}{27} sin (3 t) + \frac{t}{9} + \frac{1}{3}

y = \frac{2}{3} cos (3 t) + \frac{17}{27} sin (3 t) + \frac{t}{9} + \frac{1}{3}

What is the solution for y^{''}+9y=t+3,y(0)=1,y^'(0)=2 ?
The solution for y^{''}+9y=t+3,y(0)=1,y^'(0)=2 is y= 2/3 cos(3t)+17/27 sin(3t)+t/9+1/3