What is the solution for y^{''}-2y^'+4y=0,y^'(0)=2,y(0)=1 ?

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

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y^{''}-2y^'+4y=0,y^'(0)=2,y(0)=1

y^{^{''}} - 2 y^{^{'}} + 4 y = 0, y^{^{'}} (0) = 2, y (0) = 1

y = e^{t} (cos (3 t) + \frac{1}{3} sin (3 t))

Solution steps

y^{''} (t) - 2 y^{'} (t) + 4 y = 0

Solve linear ODE:

y = e^{t} (cos (3 t) + \frac{1}{3} sin (3 t))

y = e^{t} (cos (3 t) + \frac{1}{3} sin (3 t))

\frac{\partial}{\partial x} (- 2)

\frac{\partial}{\partial x} (y \cdot cos (y - 2 x))

n = 0 \sum \infty 9 (\frac{x - 3}{4})^{n}

\int (e^{3 x} sin (2 x)) d x

y^{^{''}} - 3 y^{^{'}} = 8 x

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