Final answer:
To solve the differential equation y’’ + 4y = sec²(2x), we first find the general solution of the associated homogeneous equation y’’ + 4y = 0, which is yh(x) = C1cos(2x) + C2sin(2x). We then apply the method of variation of parameters to find a particular solution. The general solution is the sum of yh(x) and the particular solution yp(x).
Step-by-step explanation:
The student asks to solve the differential equation y’’ + 4y = sec²(2x). To solve this second-order linear nonhomogeneous differential equation, we can use the method of variation of parameters. First, let's find the general solution of the homogeneous equation y’’ + 4y = 0.
The characteristic equation is r² + 4 = 0, which has roots r = ±2i. Thus, the general solution to the homogeneous equation is yh(x) = C1cos(2x) + C2sin(2x), where C1 and C2 are arbitrary constants.
Next, we use variation of parameters to find a particular solution yp to the nonhomogeneous equation. For this, we need to find functions u1(x) and u2(x) such that yp = u1cos(2x) + u2sin(2x).
After finding u1 and u2 by solving the system of equations derived from the variation of parameters technique, the general solution to the differential equation is y(x) = yh(x) + yp(x). It is important to note that the expressions for u1 and u2, and therefore yp, will be dependent on computing integrals which include sec²(2x).