What is the solution for y^{''}-8y^'+17y=0,y(0)=4 ?

The solution for y^{''}-8y^'+17y=0,y(0)=4 is y=e^{4t}(4cos(t)+c_{2}sin(t))

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y^{''}-8y^'+17y=0,y(0)=4

y^{^{''}} - 8 y^{^{'}} + 17 y = 0, y (0) = 4

y = e^{4 t} (4 cos (t) + c_{2} sin (t))

Solution steps

y^{''} (t) - 8 y^{'} (t) + 17 y = 0

Solve linear ODE:

y = e^{4 t} (4 cos (t) + c_{2} sin (t))

y = e^{4 t} (4 cos (t) + c_{2} sin (t))

1 - cos (θ)

x \to - 1 lim (- x^{5} - 3 x^{4})

\int 16 x - x^{2} d x

\int_{- 1}^{3} x^{5} d x

\frac{1}{x - 1}, 2

What is the solution for y^{''}-8y^'+17y=0,y(0)=4 ?
The solution for y^{''}-8y^'+17y=0,y(0)=4 is y=e^{4t}(4cos(t)+c_{2}sin(t))