What is the solution for y^{''}+4y^'+13y=52,y(0)=2,y^'(0)=1 ?

The solution for y^{''}+4y^'+13y=52,y(0)=2,y^'(0)=1 is y=e^{-2t}(-2cos(3t)-sin(3t))+4

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y^{''}+4y^'+13y=52,y(0)=2,y^'(0)=1

y^{^{''}} + 4 y^{^{'}} + 13 y = 52, y (0) = 2, y^{^{'}} (0) = 1

y = e^{- 2 t} (- 2 cos (3 t) - sin (3 t)) + 4

Solution steps

y^{''} (t) + 4 y^{'} (t) + 13 y = 52

Solve linear ODE:

y = e^{- 2 t} (- 2 cos (3 t) - sin (3 t)) + 4

y = e^{- 2 t} (- 2 cos (3 t) - sin (3 t)) + 4

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

y^{^{''}} - y^{^{'}} + y = 5 t^{2}

x^{^{''}} + 2 x^{^{'}} + 5 x = 1

q^{^{''}} + 1100 q^{^{'}} + 100000 q = 16

(D^{2} + 2 D + 1) y = 4 sin (2 x)

What is the solution for y^{''}+4y^'+13y=52,y(0)=2,y^'(0)=1 ?
The solution for y^{''}+4y^'+13y=52,y(0)=2,y^'(0)=1 is y=e^{-2t}(-2cos(3t)-sin(3t))+4