What is the solution for y^{''}+y=t^2+1 ?

The solution for y^{''}+y=t^2+1 is y=c_{1}cos(t)+c_{2}sin(t)+t^2-1

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y^{''}+y=t^2+1

y^{^{''}} + y = t^{2} + 1

y = c_{1} cos (t) + c_{2} sin (t) + t^{2} - 1

Solution steps

y^{''} (t) + y = t^{2} + 1

Solve linear ODE:

y = c_{1} cos (t) + c_{2} sin (t) + t^{2} - 1

y = c_{1} cos (t) + c_{2} sin (t) + t^{2} - 1

y^{^{''}} + 4 y^{^{'}} + 4 y = 5 e^{3 t}

y^{^{''}} + y^{^{'}} + \frac{5}{4} y = 3 cos (x)

y^{^{''}} + y^{^{'}} - 6 y = 8 + 4 e^{t}

u^{^{''}} - 3 H (t) = 7 cos (t), H (0) = - \frac{3}{4}, u^{^{'}} (0) = 0

y^{^{''}} + y^{^{'}} = 5 x + 2 e^{x}

What is the solution for y^{''}+y=t^2+1 ?
The solution for y^{''}+y=t^2+1 is y=c_{1}cos(t)+c_{2}sin(t)+t^2-1