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

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

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

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

y = t - sin (t)

Solution steps

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

Solve linear ODE:

y = t - sin (t)

y = t - sin (t)

y^{^{''}} - 9 y = 1, y (0) = 0

y^{^{''}} - 4 y^{^{'}} + 4 y = - 7 x + 4 x^{2}, y (0) = 0, y^{^{'}} (0) = 1

3 y^{^{''}} + 3 y^{^{'}} - 6 y = 4 e^{- 2 x}

y^{^{''}} + y = cos (\frac{- t + π}{2})

y^{^{''}} - y^{^{'}} - 6 y = 2 sin (3 x)

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