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

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

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

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

y = e^{t}

Solution steps

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

Solve linear ODE:

y = e^{t}

y = e^{t}

y^{^{''}} - 6 y^{^{'}} + 8 y = 0, y^{^{'}} (0) = 0

\frac{1}{8} x^{^{''}} + 16 x = 0

y^{^{'''}} - 27 y^{^{''}} + 243 y^{^{'}} - 729 y = 0

y^{^{''}} + 4 y = 0, y^{^{'}} (3 π) = 4, y (0) = 4

y^{^{''}} + 3 y^{^{'}} + α y = 0, y (0) = 1, y^{^{'}} (0) = - 8

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