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^{1+t}

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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^{1 + t}

Solution steps

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

Solve linear ODE:

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

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

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^{1+t}