What is the solution for 2y^{''}-3y^'+y=0,y(0)=0,y^'(0)=1 ?

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

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2y^{''}-3y^'+y=0,y(0)=0,y^'(0)=1

2 y^{^{''}} - 3 y^{^{'}} + y = 0, y (0) = 0, y^{^{'}} (0) = 1

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

Solution steps

2 y^{''} (t) - 3 y^{'} (t) + y = 0

Solve linear ODE:

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

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

2 (d^{4} + 11 d^{3} - 4 d^{2} - 69 d + 34) y = 0

(d^{3} - 3 d + 2) y = 0

y^{^{''}} - 2 y^{^{'}} + 5 y = 0, y (0) = 1, y^{^{'}} (0) = 3

(d^{3} + 1) y = 0

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

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