Question

In this problem you will solve the non-homogeneous differential equation y′′+9y=sec2(3x)

(1) Let C1​ and C2​ be arbitrary constants. The general solution to the related homogeneous differential equation y′′+9y=0 is the function y1​(x)=C1​y1​(x)+C2​y2​(x)=C1​ +C2​ NOTE: The order in which you enter the answers is important; that is, C1​f(x)+C2​g(x)=C1​g(x)+C2​f(x).
(2) The particular solution yp​(x) to the differential equation y′′+9y=sec2(3x) is of the form yp​(x)=y1​(x)u1​(x)+y2​(x)u2​(x) where u1​(x)= and u2​(x)=
(3) It follows that u1​(x)= and u2​(x)= ; thus yp​(x)=
(4) The most general solution to the non-homogeneous differential equation y′′+9y=sec2(3x) is y=C1​+C2​

Answer 1

The general solution to the homogeneous differential equation y''+9y=0 is y(x)=C1*cos(3x)+C2*sin(3x). For the non-homogeneous equation y''+9y=sec^2(3x), the particular solution yp(x) is obtained using the method of variation of parameters, resulting in y(x)=C1*cos(3x)+C2*sin(3x)+yp(x).

Step-by-step explanation:

The question involves solving a non-homogeneous differential equation y''+9y=sec^2(3x). First, we find the general solution to the related homogeneous equation y''+9y=0, which has the form y(x)=C1*cos(3x)+C2*sin(3x), where C1 and C2 are arbitrary constants, and cos(3x) and sin(3x) are the solutions y1(x) and y2(x) respectively. To find the particular solution yp(x) to the non-homogeneous equation, we use the method of variation of parameters. This involves finding functions u1(x) and u2(x) such that yp(x) = y1(x)u1(x) + y2(x)u2(x). After determining u1(x) and u2(x) through integration of specific formulae derived from the non-homogeneity sec^2(3x), we substitute them back to obtain yp(x). The most general solution is then the sum of the homogeneous solution and the particular solution: y(x) = C1*cos(3x) + C2*sin(3x) + yp(x).