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

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

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

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

y = e^{2 t - 2} - e^{t - 1}

Solution steps

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

Solve linear ODE:

y = e^{2 t - 2} - e^{t - 1}

y = e^{2 t - 2} - e^{t - 1}

y^{^{''''}} + 3 y^{^{'''}} + 3 y^{^{''}} + y^{^{'}} = 0

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

2 y^{^{'''}} - 4 y^{^{''}} - 2 y^{^{'}} + 4 y = 0

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 5 y = 0

81 y^{^{''}} + 72 y^{^{'}} + 16 y = 0

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