What is the solution for 4y^{''}+4y^'+y=0,y(0)=3,y^'(0)=-2 ?

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

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4y^{''}+4y^'+y=0,y(0)=3,y^'(0)=-2

4 y^{^{''}} + 4 y^{^{'}} + y = 0, y (0) = 3, y^{^{'}} (0) = - 2

y = 3 e^{- \frac{t}{2}} - \frac{e ^{- \frac{t}{2}} t}{2}

Solution steps

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

Solve linear ODE:

y = 3 e^{- \frac{t}{2}} - \frac{e ^{- \frac{t}{2}} t}{2}

y = 3 e^{- \frac{t}{2}} - \frac{e ^{- \frac{t}{2}} t}{2}

x^{^{''}} + 2 x^{^{'}} + 9 x = 0, x (0) = 0, x^{^{'}} (0) = 1

\frac{d ^{2} y}{d x ^{2}} - 1 \frac{d y}{d x} = 0

y^{^{''}} + 12 y^{^{'}} + 16 y = 0

y^{^{''}} - y - 12 y = 0

f^{^{''}} (x) + f^{^{'}} (x) = 0

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