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

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

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

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

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}

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = - 6, y^{^{'}} (0) = - 32

x^{^{''}} + 80 x = 0

y^{^{''}} - 4 y^{^{'}} + 3 y = 0, y (0) = - 2, y (0) = - 14

2 y^{^{''}} + 18 y = 0, y (0) = 2, y^{^{'}} (0) = 15

y^{^{'''''}} + 20 y^{^{'''}} = 0

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