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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} + 36 y = 0

x^{^{''}} - 2 x^{^{'}} + 5 x = 0

y^{^{''}} + 49 y = 0, y (0) = 5, y^{^{'}} (0) = - 7

y^{^{''}} + y = 0, y (π) = 0, y (0) = 0

y^{^{''}} - 13 y^{^{'}} + 42 y = 0

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