What is the solution for y^{''}-10y^'+25y=0,y(0)=2,y^'(0)=5 ?

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

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y^{''}-10y^'+25y=0,y(0)=2,y^'(0)=5

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = 2, y^{^{'}} (0) = 5

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = 2, y^{^{'}} (0) = 4

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

x^{^{''}} + 96 x = 0

y^{^{''}} - α y = 0

y^{^{''}} + 2 y^{^{'}} + y = 0, y (0) = x e^{- x} ln (x)

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