Question

find the particular integral of (D³+1)y=sin(2x+3)

Answer 1

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

Given Differential equation is

$\bf:\longmapsto\: ({D}^{3} + 1)y = sin(2x + 3)$

Now we have to find the particular integral of this Differential equation.

We know,

If Differential equation is of the form

$\rm :\longmapsto\:f(D)y = X \: then$

$\rm :\longmapsto\:P.I. \: = \: \dfrac{1}{f(D)} \: X$

Thus,

$\rm :\longmapsto\:Particular \: Integral \: is \:$

$\rm \: = \: \: \dfrac{1}{ {D}^{3} + 1} sin(2x + 3)$

$\rm \: = \: \: \dfrac{1}{ {D}^{2} \times \: D + 1} sin(2x + 3)$

Now,

$\green{\rm :\longmapsto\:Replace \: {D}^{2} = - {2}^{2} = - 4}$

$\rm \: = \: \: \dfrac{1}{ - 4 \times \: D + 1} sin(2x + 3)$

$\rm \: = \: \: \dfrac{1}{ - 4D + 1} sin(2x + 3)$

$\rm \: = \: \: \dfrac{1}{ - 4D + 1} \times \dfrac{1 + 4D}{1 + 4D} sin(2x + 3)$

$\rm \: = \: \: \dfrac{1 + 4D}{1 - {16D}^{2} } sin(2x + 3)$

$\green{\rm :\longmapsto\:Replace \: {D}^{2} = - {2}^{2} = - 4}$

$\rm \: = \: \: \dfrac{1 + 4D}{1 - 16( -4 )} sin(2x + 3)$

$\rm \: = \: \: \dfrac{1 + 4D}{1 + 64} sin(2x + 3)$

$\rm \: = \: \: \dfrac{1 + 4D}{65} sin(2x + 3)$

$\rm \: = \: \: \dfrac{1}{65}sin(2x + 3) + \dfrac{4}{65}D \: sin(2x + 3)$

$\rm \: = \: \: \dfrac{1}{65}sin(2x + 3) + \dfrac{4}{65} \: cos(2x + 3) \times 2$

$\rm \: = \: \: \dfrac{1}{65}sin(2x + 3) + \dfrac{8}{65} \: cos(2x + 3)$