Question

The solution of (D$^2$ + 16) y = cos4x is

• A. Acos4x + Bsin 4x + $\frac{x}{8}sin 4x$
• B. Acos4x + Bsin 4x - $\frac{x}{8}sin 4x$
• C. Acos4x + Bsin 4x + $\frac{x}{4}sin 4x$
• D. Acos4x + Bsin 4x - $\frac{x}{4}sin 4x$

Answer 1

Answer: (A)

Acos4x + Bsin 4x + $\frac{x}{8}sin 4x$

Answer 2

If (D$^2$ + 16)y = cos 4x

Here the auxiliary equation is m$^2$ + 16 = 0
$\Rightarrow$ m = 4

$\therefore$ Complementary function = (A cos 4x + B sin 4x) &
Particular Integral (P.I.) $= \frac{1}{D^2 + 16}. \, cos4x$

But $\frac{1}{D^2 \, + \, a^2} cos ax = \frac{x}{2a}sin ax$

$\therefore$ P.I.=$\frac{x}{2 \times 4}$. sin 4x
= $\frac{x}{8}sin4x$

$\therefore$ Solution y = Complementary function $+$ Particular Integral
$\Rightarrow$ y = A cos 4x + B sin 4x + $\frac{x}{8}$sin 4 x