What is the general solution to the second-order differential equation y” + 12y = 0?

To find the general solution of the second-order differential equation given by:

y” + 12y = 0

we begin by identifying the form of the equation. This is a linear homogeneous differential equation with constant coefficients. The standard approach is to assume a solution of the form:

y(t) = e^{rt}

where r is a constant to be determined. By substituting this assumed solution into the differential equation, we need to find the characteristic equation. First, we will differentiate:

y'(t) = r e^{rt}

y”(t) = r^2 e^{rt}

Substituting y(t), y'(t), and y”(t) into the differential equation:

r^2 e^{rt} + 12 e^{rt} = 0

Factoring out e^{rt} which is never zero gives us:

e^{rt}(r^2 + 12) = 0

This leads us to the characteristic equation:

r^2 + 12 = 0

To solve for r, we rewrite this equation:

r^2 = -12

Taking the square root of both sides, we find:

r = ±i√12 = ±2i√3

Since we have complex roots, the general solution of the differential equation will be composed of sine and cosine functions. According to the theory of homogeneous second-order differential equations, the general solution is given by:

y(t) = C_1 cos(2√3 t) + C_2 sin(2√3 t)

where C_1 and C_2 are constants determined by initial conditions or boundary conditions if provided. Thus, the general solution for the differential equation y” + 12y = 0 is:

y(t) = C_1 cos(2√3 t) + C_2 sin(2√3 t)

This solution reflects the oscillatory nature of the system described by the differential equation, characteristic of second-order linear equations with constant coefficients where the roots are purely imaginary.

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