What is the solution for y^{''}+3y=5t^4,y(0)=0,y^'(0)=0 ?

The solution for y^{''}+3y=5t^4,y(0)=0,y^'(0)=0 is y=(5t^4)/3-(20t^2)/3-40/9 cos(sqrt(3)t)+40/9

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y^{''}+3y=5t^4,y(0)=0,y^'(0)=0

y^{^{''}} + 3 y = 5 t^{4}, y (0) = 0, y^{^{'}} (0) = 0

y = \frac{5 t ^{4}}{3} - \frac{20 t ^{2}}{3} - \frac{40}{9} cos (3 t) + \frac{40}{9}

Solution steps

y^{''} (t) + 3 y = 5 t^{4}

Solve linear ODE:

y = \frac{5 t ^{4}}{3} - \frac{20 t ^{2}}{3} - \frac{40}{9} cos (3 t) + \frac{40}{9}

y = \frac{5 t ^{4}}{3} - \frac{20 t ^{2}}{3} - \frac{40}{9} cos (3 t) + \frac{40}{9}

y^{^{''}} - 10 y^{^{'}} + 24 = 0

y^{^{'''}} + 3 y^{^{''}} - 4 = 0

2 y^{^{''}} + 8 y^{^{'}} - y = 8

y^{^{''}} + 3 y^{^{'}} + 2 y = 3 e^{- 4 t}, y (0) = 1, y^{^{'}} (0) = 0

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 6 y = π x^{2} + 3

What is the solution for y^{''}+3y=5t^4,y(0)=0,y^'(0)=0 ?
The solution for y^{''}+3y=5t^4,y(0)=0,y^'(0)=0 is y=(5t^4)/3-(20t^2)/3-40/9 cos(sqrt(3)t)+40/9