What is the solution for y^{''}-8y^'+16y=0,y(0)=1,y(1)=0 ?

The solution for y^{''}-8y^'+16y=0,y(0)=1,y(1)=0 is y=e^{4t}-e^{4t}t

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y^{''}-8y^'+16y=0,y(0)=1,y(1)=0

y^{^{''}} - 8 y^{^{'}} + 16 y = 0, y (0) = 1, y (1) = 0

y = e^{4 t} - e^{4 t} t

Solution steps

y^{''} (t) - 8 y^{'} (t) + 16 y = 0

Solve linear ODE:

y = e^{4 t} - e^{4 t} t

y = e^{4 t} - e^{4 t} t

9 y^{^{''}} + 6 y^{^{'}} + 82 y = 0

y^{^{'''}} - 7 y^{^{'}} - 6 y = 0

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

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

2 \frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + 3 y = 0

What is the solution for y^{''}-8y^'+16y=0,y(0)=1,y(1)=0 ?
The solution for y^{''}-8y^'+16y=0,y(0)=1,y(1)=0 is y=e^{4t}-e^{4t}t