Answer to Question #125701 in Differential Equations for Vicky

Question

(2D^3+3D^2D'+4DD'^2+5D'^3)z=x^2+y

Answer 1

Given PDE is

$$(2D^3+3D^2D'+4DD'^2+5D'^3)z=x^2+y$$

It's a homogeneous linear diffrential equation,thus auxilary equation is

$$2m^3+3m^2+4m+5=0\\$$

Thus, $\implies m_1=\frac{1}{2} (-1 + \frac{(-63 + 2\sqrt{1086})^{1/3}}{3^{2/3}} - \frac{5}{(3 (-63 + 2\sqrt{1086}))^{1/3}}\\$

$m_2=-\frac{1}{2} -\frac{(1+i\sqrt{3})(-63 + 2\sqrt{1086})^{1/3}}{4\times3^{2/3}} +\frac{5(1-i\sqrt{3}}{4(3 (-63 + 2\sqrt{1086}))^{1/3}}\\$

$m_3=-\frac{1}{2} -\frac{(1-i\sqrt{3})(-63 + 2\sqrt{1086})^{1/3}}{4\times3^{2/3}} +\frac{5(1+i\sqrt{3}}{4(3 (-63 + 2\sqrt{1086}))^{1/3}}\\$

Hence,

$$C.F=f_1(y+m_1x)+f_2(y+m_2x)+f_3(y+m_3x)$$

Now,

$$(P.I)_1=\frac{x^2}{2D^3+3D^2D'+4DD'^2+5D'^3}\\ =\frac{1}{2D^3}\bigg(1+\bigg(\frac{3D'}{2D}+2\bigg(\frac{D'}{D}\bigg)^2+\frac{5}{2}\bigg(\frac{D'}{D}\bigg)^2\bigg)\bigg)^{-1}(x^2)\\ =\frac{1}{2D^3}(x^2)=\frac{x^5}{120}$$

Similarly,

$$(P.I)_2=\frac{y}{2D^3+3D^2D'+4DD'^2+5D'^3}\\ =\frac{1}{5D'^3}\bigg(1+\bigg(\frac{4D}{5D'}+\frac{3}{5}\bigg(\frac{D}{D'}\bigg)^2+\frac{2}{5}\bigg(\frac{D}{D'}\bigg)^2\bigg)\bigg)^{-1}(y)\\ =\frac{1}{5D'^3}(y)=\frac{y^4}{120}$$

Thus, complete solution is

$$C.F+(P.I)_1+(P.I)_2$$

which is

$$\sum_{i=1}^{3}f_i(y+m_ix)+\frac{1}{120}(x^5+y^4)$$