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

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

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

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

y = e^{t}

Solution steps

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

Solve linear ODE:

y = e^{t}

y = e^{t}

y^{^{''}} - 4 y^{^{'}} + 11 y = 0

\frac{y ^{^{''}}}{y ^{^{'}}} = 1

y^{^{''}} + 2 y^{^{'}} + y = 0, y (1) = 6, y^{^{'}} (0) = 6

y^{^{'''}} - 13 y^{^{''}} + 12 y^{^{'}} = 0

y^{^{''}} - 5 y + 6 y = 0, y (0) = - \frac{3}{5}, y^{^{'}} (0) = - \frac{13}{5}

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