What is the solution for y^{''}-2y^'+y=0,y(0)=9,y(1)=5 ?

The solution for y^{''}-2y^'+y=0,y(0)=9,y(1)=5 is y=9e^t+(5-9e)e^{t-1}t

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y^{''}-2y^'+y=0,y(0)=9,y(1)=5

y^{^{''}} - 2 y^{^{'}} + y = 0, y (0) = 9, y (1) = 5

y = 9 e^{t} + (5 - 9 e) e^{t - 1} t

Solution steps

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

Solve linear ODE:

y = 9 e^{t} + (5 - 9 e) e^{t - 1} t

y = 9 e^{t} + (5 - 9 e) e^{t - 1} t

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (0) = 2, y^{^{'}} (0) = - 13

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y^{^{'''}} - 11 y^{^{''}} + 28 y^{^{'}} = 0

(d^{4} + 2 d^{2} + 1) y = 0

\frac{d ^{3} y}{d x ^{3}} - \frac{d ^{2} y}{d x ^{2}} = 0

What is the solution for y^{''}-2y^'+y=0,y(0)=9,y(1)=5 ?
The solution for y^{''}-2y^'+y=0,y(0)=9,y(1)=5 is y=9e^t+(5-9e)e^{t-1}t