What is the solution for y^{''}-4y^'+7y=0 ?

The solution for y^{''}-4y^'+7y=0 is y=e^{2t}(c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t))

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y^{''}-4y^'+7y=0

y^{^{''}} - 4 y^{^{'}} + 7 y = 0

y = e^{2 t} (c_{1} cos (3 t) + c_{2} sin (3 t))

Solution steps

y^{''} (t) - 4 y^{'} (t) + 7 y = 0

Solve linear ODE:

y = e^{2 t} (c_{1} cos (3 t) + c_{2} sin (3 t))

y = e^{2 t} (c_{1} cos (3 t) + c_{2} sin (3 t))

\int_{- 2}^{3} \frac{28}{x ^{4}} d x

x^{2} = y^{2}, (1, - 1)

f (x) = x^{2} - 7 x + 7, at x = 0

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

x y^{^{'}} = (1 - 4 x^{2}) tan (y)

What is the solution for y^{''}-4y^'+7y=0 ?
The solution for y^{''}-4y^'+7y=0 is y=e^{2t}(c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t))