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

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

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

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))

p (θ) = 11 θ

\int x^{3} + \frac{1}{x ^{3}} d x

\frac{\partial}{\partial x} (4 z)

\int \frac{x}{2 + x ^{2}} d x

f^{^{'}} (x) = x^{\frac{2}{3}}

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