Question

Solve y"+6y'+9y=e^-3x÷e^3by using
Variation of parameters
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Answer 1

The given differential equation is

$\quad D^{2}y+6Dy+9y=\frac{e^{-3x}}{e^{3}}$

To find C.F.

Let $y=e^{mx}\:(\neq 0)$ be a trial solution of the reduced equation. Then the auxiliary equation is

$\quad m^{2}+6m+9=0$

$\Rightarrow (m+3)^{2}=0$

$\Rightarrow m=-3,\:-3$

$\therefore \color{blue}{C.F.=C_{1}e^{-3x}+C_{2}xe^{-3x}}$

To find P.I.

First we show that $e^{-3x}$ and $xe^{-3x}$ are independent solutions.

Let $y_{1}=e^{-3x},\:y_{2}=xe^{-3x}$

$\therefore W(y_{1},\:y_{2})=\left|\begin{array}{cc}e^{-3x}&xe^{-3x}\\-3e^{-3x}&e^{-3x}-3xe^{-3x}\end{array}\right|$

$\Rightarrow W(y_{1},\:y_{2})=e^{-6x}\:(\neq 0)$

$\therefore y_{1},\:y_{2}$ are linearly independent.

Let the particular solution be

$\quad y_{p}=v_{1}y_{1}+v_{2}y_{2}$,

where $y_{1}=e^{-3x},\:y_{2}=xe^{-3x}$

$\color{red}{\begin{array}{l}v_{1}=\int\frac{-y_{2}.R(x)}{W(y_{1},\:y_{2})}dx\\ \\=\int\frac{-xe^{-3x}.\frac{e^{-3x}}{e^{3}}}{e^{-6x}}dx\\ \\=-\frac{1}{e^{3}}\int x\:dx\\ \\=-\frac{1}{e^{3}}.\frac{x^{2}}{2}\\ \\=-\frac{x^{2}}{2e^{3}}\end{array}}\:\color{blue}{\begin{array}{l}v_{2}=\int\frac{y_{1}.R(x)}{W(y_{1},\:y_{2})}dx\\ \\=\int\frac{e^{-3x}.\frac{e^{-3x}}{e^{3}}}{e^{-6x}}dx\\ \\=\frac{1}{e^{3}}\int dx\\ \\=\frac{1}{e^{3}}.x\\ \\=\frac{x}{e^{3}}\end{array}}$

$\therefore y_{p}=-\frac{x^{3}}{2e^{3}}.e^{-3x}+\frac{x}{e^{3}}.xe^{-3x}$

$\Rightarrow \color{blue}{y_{p}=-\frac{1}{2}x^{3}e^{-3x-3}+x^{2}e^{-3x-3}}$

Complete solution.

$\therefore$ the complete primitive is

$\quad y=C.F.+y_{p}$

$\Rightarrow \color{red}{y=C_{1}e^{-3x}+C_{2}xe^{-3x}-\frac{1}{2}x^{3}e^{-3x-3}}\\ \color{red}{\quad\quad\quad\quad\quad\quad+x^{2}e^{-3x-3}}$