What is the solution for y^{''}+4y^'+4y=12t+16,y(0)=3,y^'(0)=-4 ?

The solution for y^{''}+4y^'+4y=12t+16,y(0)=3,y^'(0)=-4 is y=2e^{-2t}-3e^{-2t}t+3t+1

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y^{''}+4y^'+4y=12t+16,y(0)=3,y^'(0)=-4

y^{^{''}} + 4 y^{^{'}} + 4 y = 12 t + 16, y (0) = 3, y^{^{'}} (0) = - 4

y = 2 e^{- 2 t} - 3 e^{- 2 t} t + 3 t + 1

Solution steps

y^{''} (t) + 4 y^{'} (t) + 4 y = 12 t + 16

Solve linear ODE:

y = 2 e^{- 2 t} - 3 e^{- 2 t} t + 3 t + 1

y = 2 e^{- 2 t} - 3 e^{- 2 t} t + 3 t + 1

\frac{d}{d x} (\frac{2 ^{0}}{3 ^{1} + 0})

y^{^{'}} - 2 y = 0, y (0) = 2

d er i v a t i v e f (x) = cos (sin (5) x^{3})

\frac{\partial}{\partial x} (9 x^{6} y^{5} + 6 x^{8} y^{4})

y^{^{'}} + y = e^{6 t}

What is the solution for y^{''}+4y^'+4y=12t+16,y(0)=3,y^'(0)=-4 ?
The solution for y^{''}+4y^'+4y=12t+16,y(0)=3,y^'(0)=-4 is y=2e^{-2t}-3e^{-2t}t+3t+1