Question

In this problem you will solve the non-homogeneous differential equation y" + 4y = sec^2(2x) on the interval -?/4 < x < ?/4. (1) Let $C_1$ and $C_2$ be arbitrary constants. The general solution of the related homogeneous differential equation y" + 4y = 0 is the function $y_h(x) = C_1 y_1(x) + C_2 y_2(x) = C_1 \sin(2x) + C_2 \cos(2x)$. (2) The particular solution $y_p(x)$ to the differential equation y" + 4y = sec^2(2x) is of the form $y_p(x) = y_1(x) u_1(x) + y_2(x) u_2(x)$ where $u_1(x) = \frac{1}{2 \cos(2x)}$ and $u_2(x) = -\frac{1}{2} \tan(2x) \sec(2x)$. (3) It follows that $u_1(x) = \frac{1}{4} \ln(|\sec(2x) + \tan(2x)|)$ and $u_2(x) = -\frac{1}{4} \sec(2x)$; thus $y_p(x) = \boxed{}$ (4) Therefore, on the interval (-?/4, ?/4), the most general solution of the non-homogeneous differential equation y" + 4y = sec^2(2x) is y = $C_1 \cos(2x) + C_2 \sin(2x) + \boxed{}$

          In this problem you will solve the non-homogeneous differential equation
y" + 4y = sec^2(2x)
on the interval -?/4 < x < ?/4.
(1) Let $C_1$ and $C_2$ be arbitrary constants. The general solution of the related homogeneous differential equation y" + 4y = 0 is the
function $y_h(x) = C_1 y_1(x) + C_2 y_2(x) = C_1 \sin(2x) + C_2 \cos(2x)$.
(2) The particular solution $y_p(x)$ to the differential equation y" + 4y = sec^2(2x) is of the form
$y_p(x) = y_1(x) u_1(x) + y_2(x) u_2(x)$
where $u_1(x) = \frac{1}{2 \cos(2x)}$ and $u_2(x) = -\frac{1}{2} \tan(2x) \sec(2x)$.
(3) It follows that
$u_1(x) = \frac{1}{4} \ln(|\sec(2x) + \tan(2x)|)$ and $u_2(x) = -\frac{1}{4} \sec(2x)$;
thus $y_p(x) = \boxed{}$ 
(4) Therefore, on the interval (-?/4, ?/4), the most general solution of the non-homogeneous differential equation
y" + 4y = sec^2(2x)
is y = $C_1 \cos(2x) + C_2 \sin(2x) + \boxed{}$

In this problem you will solve the non-homogeneous differential equation
y" + 4y = sec^2(2x)
on the interval -?/4 < x < ?/4.
(1) Let C1 and C2 be arbitrary constants. The general solution of the related homogeneous differential equation y" + 4y = 0 is the
function yh(x) = C1 y1(x) + C2 y2(x) = C1 sin(2x) + C2 cos(2x).
(2) The particular solution yp(x) to the differential equation y" + 4y = sec^2(2x) is of the form
yp(x) = y1(x) u1(x) + y2(x) u2(x)
where u1(x) = (1)/(2 cos(2x)) and u2(x) = -(1)/(2)tan(2x) sec(2x).
(3) It follows that
u1(x) = (1)/(4)ln(|sec(2x) + tan(2x)|) and u2(x) = -(1)/(4)sec(2x);
thus yp(x) =
(4) Therefore, on the interval (-?/4, ?/4), the most general solution of the non-homogeneous differential equation
y" + 4y = sec^2(2x)
is y = C1 cos(2x) + C2 sin(2x) +

Added by Domingo T.

Question

Please give Ace some feedback