What is the solution for y^{''}+y=t^2+2,y(0)=1,y^'(0)=-1 ?

The solution for y^{''}+y=t^2+2,y(0)=1,y^'(0)=-1 is y=cos(t)-sin(t)+t^2

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y^{''}+y=t^2+2,y(0)=1,y^'(0)=-1

y^{^{''}} + y = t^{2} + 2, y (0) = 1, y^{^{'}} (0) = - 1

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 5 y^{^{'}} + 4 y = 3 x^{2}

y^{^{''}} + 4 y^{^{'}} + 3 y = 7 e^{- x}

y^{^{''}} + 8 y^{^{'}} + 16 y = e^{- 4 t}

y^{^{''}} + 9 y = sin (3 x) + e^{- x}

y^{^{''}} - y^{^{'}} - 2 y = cosh (2 x)

What is the solution for y^{''}+y=t^2+2,y(0)=1,y^'(0)=-1 ?
The solution for y^{''}+y=t^2+2,y(0)=1,y^'(0)=-1 is y=cos(t)-sin(t)+t^2