Question

OR
Solve y" +4y = cos x cos 3x.​

Answer 1

$\textbf{Given:}$

$y"+4y=cosx\;cos3x$

$y"+4y=\displaystyle\frac{1}{2}[2\;cosx\;cos3x]$

$y"+4y=\displaystyle\frac{1}{2}[cos(x+3x)+cos(x-3x)]$

$y"+4y=\displaystyle\frac{1}{2}[cos4x+cos2x]$

$\text{Characteristic equation is}$

$p^2+4=0$

$p^2=-4$

$p^2=4i^2$

$p=\pm\,2i$

$\textbf{Complementary function is}$

$e^{0x}[A\;cos2x+B\;sin2x]$

$A\;cos2x+B\;sin2x$

$\textbf{Particular integral-1}$

$=\displaystyle\frac{1}{2}(\frac{cos4x}{D^2+4})$

$=\displaystyle\frac{1}{2}(\frac{cos4x}{D^2+4})$

$=\displaystyle\frac{1}{2}(\frac{cos4x}{-16+4})$

$=\displaystyle\frac{cos4x}{-24}$

$\textbf{Particular integral-2}$

$=\displaystyle\frac{1}{2}(\frac{cos2x}{D^2+4})$

$=\displaystyle\frac{1}{2}(\frac{sin2x}{4})$

$=\displaystyle\frac{sin2x}{8}$

$\therefore\textbf{The general solution is}$

$\text{y=Complementary function+Particular integral}$

$\bf\;y=\displaystyle\;A\;cos2x+B\;sin2x-\frac{cos4x}{24}+\frac{sin2x}{8}$