What is the solution for y^{''}-4y^'+5y=0,y(0)=1,y^'(0)=0 ?

The solution for y^{''}-4y^'+5y=0,y(0)=1,y^'(0)=0 is y=e^{2t}(cos(t)-2sin(t))

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y^{''}-4y^'+5y=0,y(0)=1,y^'(0)=0

y^{^{''}} - 4 y^{^{'}} + 5 y = 0, y (0) = 1, y^{^{'}} (0) = 0

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

Solution steps

y^{''} (t) - 4 y^{'} (t) + 5 y = 0

Solve linear ODE:

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

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

y^{^{'}} = 6 x

\frac{d}{d x} (- x^{2} + 2 x + 1)

\frac{\partial}{\partial y} (e^{x + 2 y})

y^{^{'}} = x e^{x - y}

x \to 50 lim (x + 10)

What is the solution for y^{''}-4y^'+5y=0,y(0)=1,y^'(0)=0 ?
The solution for y^{''}-4y^'+5y=0,y(0)=1,y^'(0)=0 is y=e^{2t}(cos(t)-2sin(t))