Question

solving eq
y^2(x-y)p+x^2(y-x)q-z(x^2+y^2)=0

Answer 1

#Q122028

Given PDE is

y²(x-y)p+x²(y-x)=(x²+y²)z

Now we can find out its general solution by using Lagrange's method

$\implies$ dx/y²(x-y) = dy/x²(y-x) = dz/ (x²+y²)

For finding 1st pair

dx-dy/(x²+y²)(x-y) = dz/(x²+y²)

So, dx-dy/(x-y) = dz ...(1)

integrate equation 1st we get

ln(x-y)=z+lnc₁ $\implies$ ln(x-y/c₁)=z $\implies$ (x-y)/c₁=e^z $\implies$ (x-y)/e^z .....(*)

Now let us find another pair

dx+ydz/xy(x+y) = dy+xdz/xy(x+y)

$\implies$ dx+ydz = dy+xdz ....(2)

Taking integration on both side of (2 ), we get

x+yz =y+xz+c₂ $\implies$ (x-y)-z(x-y)=c₂ ......(**)

Therefore general solution is given by (*) and (**)

So, general solution is : $\Phi$ [(x-y)(1-z)]=(x-y)/e^z Ans.