Answer to Question #294720 in Differential Equations for Rohit

Question

Solve (D^3-6D^2D'+12DD'^2-8D'3)z=e^2x+y

Answer 1

Given, $(D^3-6D^2D'+12DD'^2-8D'^3)z=e^{2x+y}$

The auxiliary equation is $m^3-6m^2+12m-8=0$

The roots of the equation are m = 2 (thrice)

Therefore, the complementary function is

$C.F = f_1(y+2x)+xf_2(y+2x)+x^2f_{3}(y+2x)$

Particular integral

$\begin{aligned} P.I &= \dfrac{1}{D^3-6D^2D’+12DD’^2-8D’^3} e^{2x+y}\\ &= \dfrac{1}{2^3-6\cdot 2^2\cdot 1+12\cdot 2 \cdot 1^2-8\cdot 1^3} e^{2x+y}\\ &\text{(Replacing D = 2, D' = 1)}\\ &= \dfrac{1}{0} e^{2x+y}~~ \text{(Rule fails)}\\ &= x \dfrac{1}{3D^2 – 12DD’ +12D’^2} e^{2x+y} \\ &\text{(Differentiating with respect to D and multiplying by}~ x)\\ &= x\dfrac{1}{3 \cdot 2^2-12 \cdot 2 \cdot 1 + 12 \cdot 1^2} e^{2x+y}\\ &~~~~~~~\text{(Replacing $D = 2, D'= 1$)}\\ &= x~\frac{1}{0} e^{2x+y}~~ \text{(Rule fails)}\\ &= x^2 \dfrac{1}{6D – 12D'} e^{2x+y}\\ &\text{(Differentiating with respect to D and multiplying by $x$)}\\ &= x^2 \dfrac{1}{6 \cdot 2 – 12 \cdot 1} e^{2x+y} ~~\text{(Rule fails)}\\ &= x^3~ \dfrac{1}{6} e^{2x+y} \\ &\text{(Differentiating with respect to D and multiplying by $x$)}\\ \therefore ~P.I & = \dfrac{x^{3}}{6}e^{2x+y} \end{aligned}$

Hence the solution is given by,

$\begin{aligned} z &= C.F + P.I\\ &= f_1(y+2x)+xf_2(y+2x)+x^2f_{3}(y+2x) + \dfrac{x^{3}}{6}e^{2x+y} \end{aligned}$