What is the solution for y^{''}-10y^'+25y=0,y(1)=0,y^'(1)=-1 ?

The solution for y^{''}-10y^'+25y=0,y(1)=0,y^'(1)=-1 is y=e^{5t-5}-e^{5t-5}t

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y^{''}-10y^'+25y=0,y(1)=0,y^'(1)=-1

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (1) = 0, y^{^{'}} (1) = - 1

y = e^{5 t - 5} - e^{5 t - 5} t

Solution steps

y^{''} (t) - 10 y^{'} (t) + 25 y = 0

Solve linear ODE:

y = e^{5 t - 5} - e^{5 t - 5} t

y = e^{5 t - 5} - e^{5 t - 5} t

(D^{3} + 2 D^{2}) y = 0

2 y^{^{''}} + 5 y^{^{'}} - 3 y = 0, y (0) = 7

3 y^{^{''}} + 12 y^{^{'}} + 24 y = 0

- f^{^{''}} (t) - c f (t) = 0

6 y^{^{''}} - 7 y^{^{'}} + y = 0, y (0) = 6, y^{^{'}} (0) = \frac{1}{6}

What is the solution for y^{''}-10y^'+25y=0,y(1)=0,y^'(1)=-1 ?
The solution for y^{''}-10y^'+25y=0,y(1)=0,y^'(1)=-1 is y=e^{5t-5}-e^{5t-5}t