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^{^{'}} + 5 y = 0, y (0) = 0, y^{^{'}} (0) = 1

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

y^{^{''}} + 5 y^{^{'}} + 6 y = 0, y (0) = 1, y^{^{'}} (0) = 2

\frac{d ^{2} y}{d x ^{2}} + 6 \frac{d y}{d x} + 25 y = 0

y^{^{''}} + k^{2} 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