Question

Particular Integral of (D ^ 2 - 2D + 1) * y = e ^ x is​

Answer 1

$\large\underline{\sf{Solution-}}$

☆ Given Differential equation is

$\rm :\longmapsto\:({D}^{2} - 2D + 1)y = {e}^{x}$

So,

☆ Particular integral of Differential equation is

$\rm :\longmapsto\:P.I. = \dfrac{1}{ {D}^{2} - 2D + 1 } {e}^{x}$

$\rm \: \: = \: \dfrac{1}{1 - 2 + 1} {e}^{x} \: \: \: \: \: \: \: \: \: \: \{on \: substituting \: D = 1 \}$

$\rm \: \: = \: \dfrac{1}{0} {e}^{x} \: \: which \: is \: case \: of \: failure$

$\bf\implies \:P.I. = x \: \dfrac{1}{\dfrac{d}{dD}( {D}^{2} - 2D + 1)}{e}^{x}$

$\rm \: \: = \: x \: \dfrac{1}{2D - 2}{e}^{x}$

$\rm \: \: = \: x \: \dfrac{1}{2 \times 1 - 2}{e}^{x} \: \: \: \: \: \: \{on \: substituting \: D = 1 \}$

$\rm \: \: = \: x \: \dfrac{1}{0}{e}^{x} \: \: which \: is \: case \: of \: failure$

$\bf\implies \:P.I. = {x}^{2} \dfrac{1}{\dfrac{d}{dD}(2D - 2) } {e}^{x}$

$\rm \: \: = \: {x}^{2} \dfrac{1}{2}{e}^{x}$

$\rm \: \: = \: \dfrac{ {x}^{2} }{2}{e}^{x}$

$\bf\implies \:P.I. \: = \dfrac{ {x}^{2} }{2}{e}^{x}$