y = -2y'' + 2y' + 7y = 0
The given differential equation is a second-order linear homogeneous equation with constant coefficients. To find the general solution, we first need to find the characteristic equation, which is derived by replacing the derivatives with powers of a variable r:
2r^2 + 2r + 7 = 0
Now, we need to solve this quadratic equation for r. The quadratic formula is:
r = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 2, b = 2, and c = 7. Plugging these values into the formula, we get:
r = (-2 ± √(2^2 - 4(2)(7))) / (2(2))
r = (-2 ± √(-52)) / 4
Since the discriminant (b^2 - 4ac) is negative, we have complex roots:
r = (-2 ± √(-52)) / 4
r = (-2 ± 2i√13) / 4
r = -1/2 ± (i√13) / 2
The general solution of the differential equation is given by:
y(x) = e^(-x/2) (C1 * cos(√13 * x / 2) + C2 * sin(√13 * x / 2))
where C1 and C2 are arbitrary constants.