What is the solution for y^{''}+2y=5t^3,y(0)=0,y^'(0)=0 ?

The solution for y^{''}+2y=5t^3,y(0)=0,y^'(0)=0 is y=(5t^3)/2-(15t)/2+(15)/(2sqrt(2))sin(sqrt(2)t)

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y^{''}+2y=5t^3,y(0)=0,y^'(0)=0

y^{^{''}} + 2 y = 5 t^{3}, y (0) = 0, y^{^{'}} (0) = 0

y = \frac{5 t ^{3}}{2} - \frac{15 t}{2} + \frac{15}{2 2} sin (2 t)

Solution steps

y^{''} (t) + 2 y = 5 t^{3}

Solve linear ODE:

y = \frac{5 t ^{3}}{2} - \frac{15 t}{2} + \frac{15}{2 2} sin (2 t)

y = \frac{5 t ^{3}}{2} - \frac{15 t}{2} + \frac{15}{2 2} sin (2 t)

y^{^{''}} + 3 y^{^{'}} + 2 y = \frac{1}{( e ^{x} + 1 )}

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

2 y^{^{''}} - 2 y^{^{'}} - 4 y = 2 e^{3} t

y^{^{''}} + 2 y^{^{'}} = - e^{- 2 x}

y^{^{''}} - 4 y^{^{'}} + 5 y = 2 cos (3 x)

What is the solution for y^{''}+2y=5t^3,y(0)=0,y^'(0)=0 ?
The solution for y^{''}+2y=5t^3,y(0)=0,y^'(0)=0 is y=(5t^3)/2-(15t)/2+(15)/(2sqrt(2))sin(sqrt(2)t)