What is the solution for y^{''}-y+t^2=0 ?

The solution for y^{''}-y+t^2=0 is y=c_{1}e^t+c_{2}e^{-t}+t^2+2

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y^{''}-y+t^2=0

y^{^{''}} - y + t^{2} = 0

y = c_{1} e^{t} + c_{2} e^{- t} + t^{2} + 2

Solution steps

y^{''} (t) - y + t^{2} = 0

Solve linear ODE:

y = c_{1} e^{t} + c_{2} e^{- t} + t^{2} + 2

y = c_{1} e^{t} + c_{2} e^{- t} + t^{2} + 2

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

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

y^{^{''}} - 18 y^{^{'}} + 81 y = 480 sin (x) - 210 cos (x)

y^{^{''}} + 9 y = 2, y^{^{'}} (0) = 5

y^{^{''}} + y^{^{'}} - 12 y = sin (t)

What is the solution for y^{''}-y+t^2=0 ?
The solution for y^{''}-y+t^2=0 is y=c_{1}e^t+c_{2}e^{-t}+t^2+2