Answer to Question #164803 in Differential Equations for fahad chachar

Question

Solve the following partial differential equations 

A) (y-z)∂z/∂x +(z-x)∂z/∂y=(x-y)

B) ∂^2z/∂x^2 +7 ∂^2z/∂x∂y + 6 ∂^2z/∂y^2 =sin(x+2y)

Answer 1

(A) Given equation is-

$(y−z)p+(z−x)q=x−y$

This is Lagrange's linear equation $Pp+Qq=R$

∴ The auxiliary equation is

   

    $\dfrac{dx}{y-z}=\dfrac{dy}{z-x}=\dfrac{dz}{x-y}$

Each is equal to $\dfrac{dx+dy+dz}{y+z+z-x+x-y}$

$=\dfrac{d(x+y+z)}{0}$

d(x+y+z)=0

∴ On Integration, x+y+z=a

Also, each ratio is equal to

$\Rightarrow d(x+y+z)=0$

$\dfrac{xdx+ydy+zdz}{x(y−z)+y(z−y)+z(x−y)}$

$=\dfrac{\dfrac{1}{2}d(x^2+y^2+z^2)}{0}$

$\Rightarrow d(x^2+y^2+z^2)=0$

Integrating, $x^2+y^2+z^2=b$

∴ The general solution is

$\phi(x+y+z,x^2+y^2+z^2)=0$

(B )We have given,

$\dfrac{d^2z}{dx^2}+\dfrac{7d^2z}{dxdy}+\dfrac{6d^2z}{dy^2}=sin(x+2y)$

Then the auxiliary equation will be

$(m^2+7m+6)=0\\ m^2+m+6m+6=0\\ m(m+1)+6(m+1)=0\\ m=-1,-6$

Hence, $C.F=f1(y-x)+f2(y-6x)$

Now, Particular integral can be calculated as

$P.I= \dfrac{sin(x+2y)}{(D^2+7DD'+D'^2)}$

$= \dfrac{sin(x+2y)}{(-1)+7(-2)+6(-24)}$

$=-\dfrac{sin(x+2y)}{39}$

Therefore, the solution of partial differential equation is $C.F+P.I$

$=f1(y-x)+f2(y-6x)-\dfrac{sin(x+2y)}{39}$