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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} - 9 y = 0, y (0) = 2, y^{^{'}} (0) = 0

25 y^{^{''}} - y = 0

\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + 13 y = 0

8 y^{^{''}} + 12 y^{^{'}} + 5 y = 0

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