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

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

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

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

y = - e^{t} + 3 e^{t} t

Solution steps

y^{''} (t) - 2 y^{'} (t) + 1 y = 0

Solve linear ODE:

y = - e^{t} + 3 e^{t} t

y = - e^{t} + 3 e^{t} t

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f (t) = \frac{8 ^{8 t}}{t}

(- \frac{5}{6}, 0), (0, 6)

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