What is the solution for y^{''}-10y^'+25y=0,y(0)=1,y^'(0)=2 ?

The solution for y^{''}-10y^'+25y=0,y(0)=1,y^'(0)=2 is y=e^{5t}-3e^{5t}t

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y^{''}-10y^'+25y=0,y(0)=1,y^'(0)=2

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = 1, y^{^{'}} (0) = 2

y = e^{5 t} - 3 e^{5 t} t

Solution steps

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

Solve linear ODE:

y = e^{5 t} - 3 e^{5 t} t

y = e^{5 t} - 3 e^{5 t} t

What is the solution for y^{''}-10y^'+25y=0,y(0)=1,y^'(0)=2 ?
The solution for y^{''}-10y^'+25y=0,y(0)=1,y^'(0)=2 is y=e^{5t}-3e^{5t}t