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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

2 y^{^{'''}} - 4 y^{^{''}} - 2 y^{^{'}} + 4 y = 0

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 5 y = 0

81 y^{^{''}} + 72 y^{^{'}} + 16 y = 0

y^{^{''}} + y^{^{'}} + 1.25 y = 0

121 y^{^{''}} + 88 y^{^{'}} + 16 y = 0

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