Question

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

Answer 1

$Solution$

$(2D3+3D2D'+4DD'2+5D'3)z=x2+y$

The PDE is a homogeneous linear differential equation,thus auxilary equation is;

$$2m3+3m2+4m+5=0$$

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

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

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

$Hence,$

$C.F=f_1​(y+m_1​x)+f_2​(y+m_2​x)+f_3​(y+m_3​x)$

$Now;$

$(P.I)_1​=\frac{x^2}{2D^3+3D^2D′+4DD′^2+5D′^3​}$

$\implies \frac{1}{2D^3​}(1+(\frac{3D'}{2D}​+2(\frac{D′​}{D})^2+\frac{5}{2}​(\frac{D'}{D})^2))^{−1}(x^2)=\frac{1}{2D^3​}(x^2)=\frac{x^5​}{120}$

$For (P.I)_2;$

$(P.I)_2​=\frac{y}{2D^3+3D^2D′+4DD′^2+5D′^3​}$

$\implies \frac{1}{5D'^3​}(1+(\frac{4D}{5D'}​+\frac{3}{5}(\frac{D​}{D'})^2+\frac{2}{5}​(\frac{D}{D'})^2))^{−1}(y)=\frac{1}{5D'^3​}(y)=\frac{y^4}{120}$

$Hence,$

$$C.F+(P.I)_1​+(P.I)_2​$$

$\implies ∑_{i=1}^{3}​f_i​(y+m_i​x)+\frac{1}{120}​(x^5+y^4)$