Subject : Applied Mathematics 2

Topic : Linear Differential Equations With Constant Coefficients and Variable Coefficients Of Higher Order

Difficulty: High

Solution:

$x^{2} \frac{d^{2} y}{d x^{2}}+3 x \frac{d y}{d x}+3 y=\frac{\log x \cdot \cos (\log x)}{x}$

Putting $z=\log x$ and $x=e^{z},$ we get

$[D(D-1)+3 D+3] y=e^{-z} \cdot z \cdot \cos z$

$\left[D^{2}+2 D+3\right] y=e^{-z} \cdot z \cdot \cos z$

$\therefore$ The A.E. is $D^{2}+2 D+3=0$

$\therefore D=\frac{-2 \pm 2 \sqrt{2} . i}{2}=-1 \pm \sqrt{2} . i$

$\therefore$ The C.F. is $y=e^{-z}\left(C_{1} \cos \sqrt{2} z+C_{2} \sin \sqrt{2} z\right)$

P.I. $=\frac{1}{D^{2}+2 D+3} e^{-z} \cdot z \cdot \cos z$

$\quad=e^{-z} \cdot \frac{1}{(D-1)^{2}+2(D-1)+3} \cdot z \cdot \cos z=e^{-z} \cdot \frac{1}{D^{2}+2} \cdot z \cdot \cos z$

$=e^{-z}\left[z-\frac{1}{D^{2}+2} .2 D\right] \cdot \frac{1}{D^{2}+2} \cdot \cos z$

$=e^{-z}\left[z-\frac{1}{D^{2}+2} \cdot 2 D\right] \cos z=e^{-z}\left[z \cos z+\frac{1}{D^{2}+2} \cdot 2 \sin z\right]$

$=e^{-z}[z \cos z+2 \sin z]$

The complete solution is,

$y=C . F .+P .I$

$y=e^{-z}\left(C_{1} \cos \sqrt{2} z+C_{2} \sin \sqrt{2} z\right)+e^{-z}[z \cos z+2 \sin z]$

$y=\frac{1}{x}\left(c_{1} \cos \sqrt{2} \log x+C_{2} \sin \sqrt{2} \log x\right)+\frac{1}{x}[\log x \cos \log x+2 \sin \log x ]$