Question

solve (D^3+3D^2+2D)y=x^2​

Answer 1

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

Given Differential equation is

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

Its Auxiliary equation is

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

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

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

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

$\rm :\implies\:D = 0, - 1, - 2$

Therefore,

Complementary function is given by

$\rm :\longmapsto\:y_h = a \: + \: b {e}^{ - x} + {ce}^{ - 2x} - - - (1)$

Now, we have to find particular solution to non - homogeneous solution using method of undetermined coefficients.

Let

$\rm :\longmapsto\:y_p = x(p {x}^{2} + qx + r) - - (2)$

On differentiating thrice, w. r. t. x, we get

$\rm :\longmapsto\:y_p =p {x}^{3} + q {x}^{2} + rx$

$\rm :\longmapsto\:y_p' = 3p {x}^{2} + 2qx + r$

$\rm :\longmapsto\:y_p'' = 6p {x} + 2q$

$\rm :\longmapsto\:y_p''' = 6p$

Now, given differential equation is

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

$\rm :\longmapsto\:{D}^{3}y + {3D}^{2}y + 2Dy = {x}^{2}$

$\rm :\longmapsto\:y_p''' + 3y_p'' + 2y_p' = {x}^{2}$

$\rm :\longmapsto\:6p + 3(6px + 2q) + 2( {3px}^{2} + 2qx + r) = {x}^{2}$

$\rm :\longmapsto\:6p + 18px + 6q+ {6px}^{2} + 4qx + 2r= {x}^{2}$

$\rm :\longmapsto\: {6px}^{2} + (18p + 4q)x + 6p + 6q + 2r = {x}^{2}$

Now,

On comparing the coefficients, we get

$\rm :\longmapsto\:6p = 1\rm \implies\:p = \dfrac{1}{6}$

$\rm :\longmapsto\:18p + 4q = 0\rm \implies\:3 + 4q = 0\rm \implies\:q = - \dfrac{3}{4}$

and

$\rm :\longmapsto\:6p + 6q + 2r = 0$

$\rm \implies\:r = - \dfrac{7}{4}$

On substituting the values of p, q and r, in equation (2),

$\rm :\longmapsto\:y_p = x \bigg( \dfrac{1}{6} {x}^{2} - \dfrac{3}{4} x - \dfrac{7}{4} \bigg )$

Hence,

General solution is given by

$\rm :\longmapsto\:y = y_h + y_p$

$\rm \implies\:y = a +b {e}^{ - x} + {ce}^{ - 2x} + x \bigg( \dfrac{1}{6} {x}^{2} - \dfrac{3}{4} x - \dfrac{7}{4} \bigg )$