What is the solution for 4/32 x^{''}+x^'+2x=0,x(0)=-1,x^'(0)=16 ?

The solution for 4/32 x^{''}+x^'+2x=0,x(0)=-1,x^'(0)=16 is x=-e^{-4t}+12e^{-4t}t

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4/32 x^{''}+x^'+2x=0,x(0)=-1,x^'(0)=16

\frac{4}{32} x^{^{''}} + x^{^{'}} + 2 x = 0, x (0) = - 1, x^{^{'}} (0) = 16

x = - e^{- 4 t} + 12 e^{- 4 t} t

Solution steps

\frac{4}{32} x^{''} (t) + x^{'} (t) + 2 x = 0

Solve linear ODE:

x = - e^{- 4 t} + 12 e^{- 4 t} t

x = - e^{- 4 t} + 12 e^{- 4 t} t

\frac{d ^{2} y}{d t ^{2}} - 5 \frac{d y}{d t} + 6 y = 0, y (1) = 3

x^{^{''}} - 3 x^{^{'}} = 0

y^{^{''}} + 36 y = 0, y (0) = 5, y^{^{'}} (0) = - 12

y^{^{''}} + 4 \cdot y^{^{'}} + 4 \cdot y = 0

y^{^{''}} + 12 y^{^{'}} + 36 y = 0, y (0) = - 1, y^{^{'}} (0) = 9

What is the solution for 4/32 x^{''}+x^'+2x=0,x(0)=-1,x^'(0)=16 ?
The solution for 4/32 x^{''}+x^'+2x=0,x(0)=-1,x^'(0)=16 is x=-e^{-4t}+12e^{-4t}t