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Question Number 91384 by jagoll last updated on 30/Apr/20
$${particular}\:{integral}\: \\ $$$${y}''+\mathrm{3}{y}'+\mathrm{2}{y}\:=\:\mathrm{4cos}\:^{\mathrm{2}} {x} \\ $$
Commented by jagoll last updated on 30/Apr/20
![PI ⇒y_p = ((4cos^2 x)/(D^2 +3D+2)) = ((4cos^2 x)/(2D+3)) = (((2D−3)(4cos^2 x))/(−9)) = −(1/9)[ 8(−sin 2x)−12cos^2 x ] = −(1/9)(−8sin 2x−6−6cos 2x) = ((8sin 2x)/9)+((6cos 2x)/9)+(6/9)](Q91386.png)
$${PI}\:\Rightarrow{y}_{{p}} \:=\:\frac{\mathrm{4cos}\:^{\mathrm{2}} {x}}{{D}^{\mathrm{2}} +\mathrm{3}{D}+\mathrm{2}} \\ $$$$=\:\frac{\mathrm{4cos}\:^{\mathrm{2}} {x}}{\mathrm{2}{D}+\mathrm{3}}\:=\:\frac{\left(\mathrm{2}{D}−\mathrm{3}\right)\left(\mathrm{4cos}\:^{\mathrm{2}} {x}\right)}{−\mathrm{9}} \\ $$$$=\:−\frac{\mathrm{1}}{\mathrm{9}}\left[\:\mathrm{8}\left(−\mathrm{sin}\:\mathrm{2}{x}\right)−\mathrm{12cos}\:^{\mathrm{2}} {x}\:\right] \\ $$$$=\:−\frac{\mathrm{1}}{\mathrm{9}}\left(−\mathrm{8sin}\:\mathrm{2}{x}−\mathrm{6}−\mathrm{6cos}\:\mathrm{2}{x}\right) \\ $$$$=\:\frac{\mathrm{8sin}\:\mathrm{2}{x}}{\mathrm{9}}+\frac{\mathrm{6cos}\:\mathrm{2}{x}}{\mathrm{9}}+\frac{\mathrm{6}}{\mathrm{9}} \\ $$
Commented by jagoll last updated on 30/Apr/20
$${it}\:{correct}\:? \\ $$
Commented by 675480065 last updated on 30/Apr/20
$$\mathrm{ok}\:\mathrm{sir} \\ $$
Commented by jagoll last updated on 30/Apr/20
$${are}\:{sure}\:? \\ $$$$ \\ $$