What is the solution for 2y^{''}+16y^'+32y=0,y(0)=-3,y^'(0)=10 ?

The solution for 2y^{''}+16y^'+32y=0,y(0)=-3,y^'(0)=10 is y=-3e^{-4t}-2e^{-4t}t

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2y^{''}+16y^'+32y=0,y(0)=-3,y^'(0)=10

2 y^{^{''}} + 16 y^{^{'}} + 32 y = 0, y (0) = - 3, y^{^{'}} (0) = 10

y = - 3 e^{- 4 t} - 2 e^{- 4 t} t

Solution steps

2 y^{''} (t) + 16 y^{'} (t) + 32 y = 0

Solve linear ODE:

y = - 3 e^{- 4 t} - 2 e^{- 4 t} t

y = - 3 e^{- 4 t} - 2 e^{- 4 t} t

\frac{d}{d x} (f (x) (x)^{3} (x))

f (x) = cos (2 x)

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

\int sec^{5} (x) tan^{3} (x) d x

x \to 0 lim (\frac{( x + 1 ) ^{5} - 1}{x})

What is the solution for 2y^{''}+16y^'+32y=0,y(0)=-3,y^'(0)=10 ?
The solution for 2y^{''}+16y^'+32y=0,y(0)=-3,y^'(0)=10 is y=-3e^{-4t}-2e^{-4t}t