Subject : Applied Mathematics 2

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

Difficulty: High

Solution:

$x^{2} \frac{d^{3} y}{d x^{3}}+3 x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}+\frac{y}{x}=4 \log x$

The given diff. eqn is Cauchy's homogeneous eqn.

Multiply the given eqn by x,

$x^{3} \frac{d^{3} y}{d x^{3}}+3 x^{2} \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}+y=4 x \log x$

Put $x=e^{z} \quad \log x=z$

Diff. W.r.t x ,

$\frac{1}{x}=\frac{d z}{d x} \quad$ but $\frac{d y}{d x}=\frac{d y}{d z} \cdot \frac{d z}{d x}=\frac{d y}{d z} \frac{1}{x}$

$\begin{aligned} \therefore \mathbf{x} \frac{d y}{d x} &=D y \\ x^{2} \frac{d^{2} y}{d x^{2}} &=D(D-1) y \\ x^{3} \frac{d^{3} y}{d x^{3}} &= D(D-1)(D-2) y \quad where D=\frac{d}{d z} \\ \therefore \quad & [\mathrm{D}(\mathrm{D}-1)(\mathrm{D}-2)+3 \mathrm{D}(\mathrm{D}-1)+\mathrm{D}+1] \mathrm{y}=4 \mathrm{z} . e^{z} \\ \therefore \quad & \left[D^{3}+1\right] \mathrm{y}=4 \mathrm{z} . e^{z} \end{aligned}$

For complementary solution,

$f(D)=0$

$\therefore\left[D^{3}+1\right]=0$

Roots are: $\mathrm{D}=-1, \frac{1}{2}+i \frac{\sqrt{3}}{2}, \frac{1}{2}-i \frac{\sqrt{3}}{2}$

Roots of the eqn are real and complex.

$ \mathbf{ \therefore y_{c}=c_{1} e^{-z}+e^{z / 2}\left(c_{2} \cos \frac{\sqrt{3} z}{2}+c_{3} \sin \frac{\sqrt{3} z}{2}\right) }$

For particular integral,

$\begin{aligned} y_{p} &=\frac{1}{f(D)} X=\frac{1}{\left(D^{3}+1\right)} 4 z \cdot e^{z} \\ &=4 e^{z} \frac{1}{(D+1)^{3}+1} \mathrm{Z} \\ &=4 e^{z} \frac{1}{D^{3}+3 D^{2}+3 D+2} Z \\ \therefore y_{p} &= \mathbf{ e^{z}(2 z-3) } \end{aligned}$

The general solution of given diff. eqn is

$\quad y_{g}=y_{c}+y_{p}=c_{1} e^{-z}+e^{z / 2}\left(c_{2} \cos \frac{\sqrt{3} z}{2}+c_{3} \sin \frac{\sqrt{3} z}{2}\right)+e^{z}(2 z-3)$

Resubstitute z

$ \mathbf{ \therefore y_{g}=\frac{c_{1}}{x}+\sqrt{x}\left(c_{2} \cos \frac{\sqrt{3} \log x}{2}+c_{3} \sin \frac{\sqrt{3} \log x}{2}\right)+x(2 \log x-3) }$