What is the solution for y^{''}+4y=4t,y(0)=1,y^'(0)=5 ?

The solution for y^{''}+4y=4t,y(0)=1,y^'(0)=5 is y=cos(2t)+2sin(2t)+t

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y^{''}+4y=4t,y(0)=1,y^'(0)=5

y^{^{''}} + 4 y = 4 t, y (0) = 1, y^{^{'}} (0) = 5

y = cos (2 t) + 2 sin (2 t) + t

Solution steps

y^{''} (t) + 4 y = 4 t

Solve linear ODE:

y = cos (2 t) + 2 sin (2 t) + t

y = cos (2 t) + 2 sin (2 t) + t

y^{^{''}} + 4 y = 9 t

y^{^{''}} + 4 y = 3 csc (x)

y^{^{'''}} - 6 y^{^{''}} + 11 y^{^{'}} - 6 y = 2 x e^{- x}

y^{^{''}} + 4 y = cos^{2} (2 x)

y^{^{''}} - 5 y^{^{'}} + 6 y = (x + 1) e^{- x}

What is the solution for y^{''}+4y=4t,y(0)=1,y^'(0)=5 ?
The solution for y^{''}+4y=4t,y(0)=1,y^'(0)=5 is y=cos(2t)+2sin(2t)+t