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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 7 y^{^{'}} + 10 y = 0, y^{^{'}} (0) = - \frac{1}{3}

(D^{2} - 6 D + 25) y = 0

4 y^{^{''}} + 4 y^{^{'}} + y = 0, y (0) = - 1, y^{^{'}} (0) = 2

\frac{d ^{2} y}{d x ^{2}} + 36 y = 0

x^{^{''}} - 2 x^{^{'}} + 5 x = 0

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