Question

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

• A. Acos4x + Bsin 4x + $\frac{x}{8}sin4x$
• B. Acos4x + Bsin 4x - $\frac{x}{8}sin4x$
• C. Acos4x + Bsin 4x + $\frac{x}{4}sin4x$
• D. Acos4x + Bsin 4x - $\frac{x}{4}sin4x$

Answer 1

Answer: (A)

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 $\frac{1}{D^2+a^2}cosax=\frac{x}{2a}sinax$
$\therefore P.I.=\frac{x}{2\times 4}.sin4x=\frac{x}{8}sin4x$
$\therefore$ Solution y = Complementary function
+ Particular Integral
$\Rightarrow$ y = A cos 4x + B sin 4x + $\frac{x}{8}$sin 4 x