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

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

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

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

y = - e^{t} + (2 + e) e^{t - 1} t

Solution steps

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

Solve linear ODE:

y = - e^{t} + (2 + e) e^{t - 1} t

y = - e^{t} + (2 + e) e^{t - 1} t

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

(D^{5} - D^{3}) y = 0

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

y^{^{''}} - 9 y^{^{'}} + 20 y = 0, y (0) = 3, y^{^{'}} (0) = 2

y^{^{''}} + \frac{1}{100} y^{^{'}} = 0

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