Question

d^2y/dx^2+3dy/dx+2y=4cos^2x

Answer 1

Answer:

Symbolic form of the given equation, (D2 + 3D + 2)y = eex Auxiliary equation becomes D2 + 3D + 2 = 0 or D = –1, –2. C.F. = c1e– x + c2e– 2x

Answer 2

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

Given Differential equation is

$\rm :\longmapsto\:\dfrac{ {d}^{2} y}{ {dx}^{2} } + 3\dfrac{dy}{dx} + 2y \: = \: {4cos}^{2}x$

The symbolic form of the equation is

$\rm :\longmapsto\: ({D}^{2} + 3D + 2)y = {4cos}^{2}x$

So, Auxiliary equation is

$\rm :\longmapsto\: {D}^{2} + 3D + 2 = 0$

$\rm :\longmapsto\: {D}^{2} + 2D + D+ 2 = 0$

$\rm :\longmapsto\:D(D + 2) + 1(D + 2) = 0$

$\rm :\longmapsto\:(D + 2)(D + 1) = 0$

$\bf\implies \:D = - 1 \: \: or \: \: D = - 2$

So,

$\boxed{ \bf{ \:C F. = a {e}^{ - x} + b {e}^{ - 2x}}}$

To evaluate Particular Integral, P. I.

$\rm :\longmapsto\:P. I. = \dfrac{1}{ {D}^{2} + 3D + 2} {4cos}^{2}x$

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

$\rm \:  =  \:  2\: \dfrac{1}{ {D}^{2} + 3D + 2} ( 1 + cos2x)$

$\rm \:  =  \:  2\: \dfrac{1}{ {D}^{2} + 3D + 2}(1) + \dfrac{1}{ {D}^{2} + 3D + 2}(cos2x)$

$\rm \:  =  \:  2\: \dfrac{1}{ {1}^{2} + 3(1) + 2} + \dfrac{1}{ { - 2}^{2} + 3D + 2}(cos2x)$

$\red{\bigg \{ \because \: {D}^{2} = - {2}^{2} \bigg \}}$

$\rm \:  =  \: \: \dfrac{2}{6} + \dfrac{1}{ - 4 + 3D + 2}(cos2x)$

$\rm \:  =  \: \: \dfrac{1}{3} + \dfrac{1}{ 3D - 2}(cos2x)$

$\rm \:  =  \: \: \dfrac{1}{3} + \dfrac{3D + 2}{ (3D - 2)(3D + 2)}(cos2x)$

$\rm \:  =  \: \: \dfrac{1}{3} + \dfrac{3D + 2}{ {9D}^{2} - 4}(cos2x)$

$\rm \:  =  \: \: \dfrac{1}{3} + \dfrac{3D + 2}{ - 9{(2)}^{2} - 4}(cos2x)$

$\rm \:  =  \: \: \dfrac{1}{3} + \dfrac{3D + 2}{ - 36 - 4}(cos2x)$

$\rm \:  =  \: \: \dfrac{1}{3} + \dfrac{3D(cos2x) + 2cos2x}{ -40}$

$\rm \:  =  \: \: \dfrac{1}{3} + \dfrac{3( - sin2x \times 2)+ 2cos2x}{ -40}$

$\rm \:  =  \: \: \dfrac{1}{3} + \dfrac{ - 3sin2x+ cos2x}{ -20}$

$\rm \:  =  \: \: \dfrac{1}{3} + \dfrac{ 3sin2x - cos2x}{20}$

Hence, The complete solution is

$\boxed{ \sf{ \:y = C F. + P. I. }}$

$\rm \: y =  \: \: a {e}^{ - x} + b {e}^{ - 2x} + \dfrac{1}{3} + \dfrac{ 3sin2x - cos2x}{20}$