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

The solution for 8y^{''}+y=t,y(0)=0,y^'(0)=0 is y=t-2sqrt(2)sin(t/(2sqrt(2)))

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

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

y = t - 22 sin (\frac{t}{2 2})

Solution steps

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

Solve linear ODE:

y = t - 22 sin (\frac{t}{2 2})

y = t - 22 sin (\frac{t}{2 2})

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

y^{^{''}} + y^{^{'}} + \frac{256}{3} x = 50 cos (3 x)

y^{^{''}} - y^{^{'}} + y = 25 sin (x)

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

y^{^{''}} + 2 y^{^{'}} = e^{t}

What is the solution for 8y^{''}+y=t,y(0)=0,y^'(0)=0 ?
The solution for 8y^{''}+y=t,y(0)=0,y^'(0)=0 is y=t-2sqrt(2)sin(t/(2sqrt(2)))