What is the solution for y^{''}-10y^'+25y=0,y(0)=-6,y^'(0)=-32 ?

The solution for y^{''}-10y^'+25y=0,y(0)=-6,y^'(0)=-32 is y=-6e^{5t}-2e^{5t}t

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y^{''}-10y^'+25y=0,y(0)=-6,y^'(0)=-32

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = - 6, y^{^{'}} (0) = - 32

y = - 6 e^{5 t} - 2 e^{5 t} t

Solution steps

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

Solve linear ODE:

y = - 6 e^{5 t} - 2 e^{5 t} t

y = - 6 e^{5 t} - 2 e^{5 t} t

x^{^{''}} + 80 x = 0

y^{^{''}} - 4 y^{^{'}} + 3 y = 0, y (0) = - 2, y (0) = - 14

2 y^{^{''}} + 18 y = 0, y (0) = 2, y^{^{'}} (0) = 15

y^{^{'''''}} + 20 y^{^{'''}} = 0

\frac{5}{8} x^{^{''}} + 40 x = 0, x (0) = \frac{1}{3}, x^{^{'}} (0) = 0

What is the solution for y^{''}-10y^'+25y=0,y(0)=-6,y^'(0)=-32 ?
The solution for y^{''}-10y^'+25y=0,y(0)=-6,y^'(0)=-32 is y=-6e^{5t}-2e^{5t}t