What is the solution for y^{''}+8y=2t^4,y(0)=0,y^'(0)=0 ?

The solution for y^{''}+8y=2t^4,y(0)=0,y^'(0)=0 is y=(t^4)/4-(3t^2)/8-(3cos(2^{1/2+1}t))/(32)+3/32

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y^{''}+8y=2t^4,y(0)=0,y^'(0)=0

y^{^{''}} + 8 y = 2 t^{4}, y (0) = 0, y^{^{'}} (0) = 0

y = \frac{t ^{4}}{4} - \frac{3 t ^{2}}{8} - \frac{3 cos ( 2 ^{\frac{1}{2} + 1} t )}{32} + \frac{3}{32}

Solution steps

y^{''} (t) + 8 y = 2 t^{4}

Solve linear ODE:

y = \frac{t ^{4}}{4} - \frac{3 t ^{2}}{8} - \frac{3 cos ( 2 ^{\frac{1}{2} + 1} t )}{32} + \frac{3}{32}

y = \frac{t ^{4}}{4} - \frac{3 t ^{2}}{8} - \frac{3 cos ( 2 ^{\frac{1}{2} + 1} t )}{32} + \frac{3}{32}

y^{^{''}} - 4 y^{^{'}} + 4 y = x^{- 1} e^{2 x}

16 y^{^{''}} - 24 y^{^{'}} + 9 y = 5 - 3 x

y^{^{''}} + 4 y^{^{'}} + 4 y = e^{- 2 x} \cdot x^{- 2}

y^{^{''}} + 4 y^{^{'}} + 40 y = 50 sin (6 t)

y^{^{''}} - 4 y + 3 y = 2 x + 1

What is the solution for y^{''}+8y=2t^4,y(0)=0,y^'(0)=0 ?
The solution for y^{''}+8y=2t^4,y(0)=0,y^'(0)=0 is y=(t^4)/4-(3t^2)/8-(3cos(2^{1/2+1}t))/(32)+3/32