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)-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) - 3 sin (3 t))

Solution steps

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

Solve linear ODE:

y = e^{t} (cos (3 t) - 3 sin (3 t))

y = e^{t} (cos (3 t) - 3 sin (3 t))

y^{^{''}} - 4 y^{^{'}} + 4 y = 0, y (0) = 5, y^{^{'}} = 2

y^{^{''}} + 16 y^{^{'}} + 64 y = 0, y (0) = 5, y^{^{'}} (0) = 0

x^{^{''}} - 16 x = 0

y^{^{''}} - 18 y^{^{'}} + 81 y = 0, y (0) = 0, y^{^{'}} (0) = 5

\frac{d ^{2} y}{d x ^{2}} - 13 \frac{d y}{d x} + 40 y = 0

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)-sqrt(3)sin(sqrt(3)t))