What is the solution for y^{''}-6y^'+25y=0,y(0)=2,y^'(0)=2 ?

The solution for y^{''}-6y^'+25y=0,y(0)=2,y^'(0)=2 is y=e^{3t}(2cos(4t)-sin(4t))

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y^{''}-6y^'+25y=0,y(0)=2,y^'(0)=2

y^{^{''}} - 6 y^{^{'}} + 25 y = 0, y (0) = 2, y^{^{'}} (0) = 2

y = e^{3 t} (2 cos (4 t) - sin (4 t))

Solution steps

y^{''} (t) - 6 y^{'} (t) + 25 y = 0

Solve linear ODE:

y = e^{3 t} (2 cos (4 t) - sin (4 t))

y = e^{3 t} (2 cos (4 t) - sin (4 t))

What is the solution for y^{''}-6y^'+25y=0,y(0)=2,y^'(0)=2 ?
The solution for y^{''}-6y^'+25y=0,y(0)=2,y^'(0)=2 is y=e^{3t}(2cos(4t)-sin(4t))