What is the solution for y^{''}-14y^'+49y=0,y(0)=1,y(1)=0 ?

The solution for y^{''}-14y^'+49y=0,y(0)=1,y(1)=0 is y=e^{7t}-e^{7t}t

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y^{''}-14y^'+49y=0,y(0)=1,y(1)=0

y^{^{''}} - 14 y^{^{'}} + 49 y = 0, y (0) = 1, y (1) = 0

y = e^{7 t} - e^{7 t} t

Solution steps

y^{''} (t) - 14 y^{'} (t) + 49 y = 0

Solve linear ODE:

y = e^{7 t} - e^{7 t} t

y = e^{7 t} - e^{7 t} t

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

y^{^{'''}} - 4 y^{^{''}} - 32 y^{^{'}} = 0

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

y^{^{''}} + 2 y^{^{'}} + 17 y = 0, y^{^{'}} (0) = - 2

y^{^{''}} + \frac{10}{3} y^{^{'}} + a y = 0, y (0) = 1, y^{^{'}} (0) = \frac{5}{3}

What is the solution for y^{''}-14y^'+49y=0,y(0)=1,y(1)=0 ?
The solution for y^{''}-14y^'+49y=0,y(0)=1,y(1)=0 is y=e^{7t}-e^{7t}t