What is the solution for y^{''}-4y^'+9y=0,y(0)=0,y^'(0)=-8 ?

The solution for y^{''}-4y^'+9y=0,y(0)=0,y^'(0)=-8 is y=-8/(sqrt(5))e^{2t}sin(sqrt(5)t)

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y^{''}-4y^'+9y=0,y(0)=0,y^'(0)=-8

y^{^{''}} - 4 y^{^{'}} + 9 y = 0, y (0) = 0, y^{^{'}} (0) = - 8

y = - \frac{8}{5} e^{2 t} sin (5 t)

Solution steps

y^{''} (t) - 4 y^{'} (t) + 9 y = 0

Solve linear ODE:

y = - \frac{8}{5} e^{2 t} sin (5 t)

y = - \frac{8}{5} e^{2 t} sin (5 t)

f (x) = \frac{x ^{3}}{9}

\frac{d y}{d x} - 3 y = - 7 sin (x) - 7 cos (x)

x \to + 1 lim (e^{x - 1} - 1)

cos (x) + x sin (x)

\frac{\partial}{\partial x} (\frac{1}{3} x^{3} y^{3})

What is the solution for y^{''}-4y^'+9y=0,y(0)=0,y^'(0)=-8 ?
The solution for y^{''}-4y^'+9y=0,y(0)=0,y^'(0)=-8 is y=-8/(sqrt(5))e^{2t}sin(sqrt(5)t)