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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

2 y^{^{''}} + 12 y^{^{'}} + 68 y = 0

y^{^{''}} + 14 y = 0

34 y + 10 y^{^{'}} + y^{^{''}} = 0, y (0) = - 3, y^{^{'}} (0) = 6

y^{^{''}} - 25 y = 0, y (0) = e^{5 x}

y^{^{''}} + 24 y = 0

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