What is the solution for y^{''}+6y=t^4,y(0)=0,y^'(0)=0 ?

The solution for y^{''}+6y=t^4,y(0)=0,y^'(0)=0 is y=(t^4)/6-(t^2)/3-1/9 cos(sqrt(6)t)+1/9

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y^{''}+6y=t^4,y(0)=0,y^'(0)=0

y^{^{''}} + 6 y = t^{4}, y (0) = 0, y^{^{'}} (0) = 0

y = \frac{t ^{4}}{6} - \frac{t ^{2}}{3} - \frac{1}{9} cos (6 t) + \frac{1}{9}

Solution steps

y^{''} (t) + 6 y = t^{4}

Solve linear ODE:

y = \frac{t ^{4}}{6} - \frac{t ^{2}}{3} - \frac{1}{9} cos (6 t) + \frac{1}{9}

y = \frac{t ^{4}}{6} - \frac{t ^{2}}{3} - \frac{1}{9} cos (6 t) + \frac{1}{9}

What is the solution for y^{''}+6y=t^4,y(0)=0,y^'(0)=0 ?
The solution for y^{''}+6y=t^4,y(0)=0,y^'(0)=0 is y=(t^4)/6-(t^2)/3-1/9 cos(sqrt(6)t)+1/9