What is the solution for y^{''}-6y^'+25y=0,y(0)=1,y^'(0)=-11 ?

The solution for y^{''}-6y^'+25y=0,y(0)=1,y^'(0)=-11 is y=e^{3t}(cos(4t)-7/2 sin(4t))

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y^{''}-6y^'+25y=0,y(0)=1,y^'(0)=-11

y^{^{''}} - 6 y^{^{'}} + 25 y = 0, y (0) = 1, y^{^{'}} (0) = - 11

y = e^{3 t} (cos (4 t) - \frac{7}{2} sin (4 t))

Solution steps

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

Solve linear ODE:

y = e^{3 t} (cos (4 t) - \frac{7}{2} sin (4 t))

y = e^{3 t} (cos (4 t) - \frac{7}{2} sin (4 t))

What is the solution for y^{''}-6y^'+25y=0,y(0)=1,y^'(0)=-11 ?
The solution for y^{''}-6y^'+25y=0,y(0)=1,y^'(0)=-11 is y=e^{3t}(cos(4t)-7/2 sin(4t))