What is the solution for y^{''}-8y^'+80y=0,y(0)=10,y^'(0)=8 ?

The solution for y^{''}-8y^'+80y=0,y(0)=10,y^'(0)=8 is y=e^{4t}(10cos(8t)-4sin(8t))

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y^{''}-8y^'+80y=0,y(0)=10,y^'(0)=8

y^{^{''}} - 8 y^{^{'}} + 80 y = 0, y (0) = 10, y^{^{'}} (0) = 8

y = e^{4 t} (10 cos (8 t) - 4 sin (8 t))

Solution steps

y^{''} (t) - 8 y^{'} (t) + 80 y = 0

Solve linear ODE:

y = e^{4 t} (10 cos (8 t) - 4 sin (8 t))

y = e^{4 t} (10 cos (8 t) - 4 sin (8 t))

4 y^{^{'''}} - 3 y^{^{''}} - 25 y^{^{'}} - 6 y = 0

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

4 \frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + y = 0

41 y - 8 y^{^{'}} + y^{^{''}} = 0, y (0) = - 3, y^{^{'}} (0) = - 32

y^{^{''}} + 4 y = 0, y (0) = 3, y^{^{'}} (0) = 6

What is the solution for y^{''}-8y^'+80y=0,y(0)=10,y^'(0)=8 ?
The solution for y^{''}-8y^'+80y=0,y(0)=10,y^'(0)=8 is y=e^{4t}(10cos(8t)-4sin(8t))