Question

solve the partial diffrential equation y^z/x.p+xzq=y^2​

Answer 1

Answer:

Step-by-step explanation:

The subsidiary equations are -

xdx/y²z = dy/xz = dz/y²

Taking the first two ratios,

xdx/y²z = dy/xz

⇒xdx/y² = dy/x

⇒x²dx = y²dy

Integrating we get,

x³-y³ = c1

Similarly, from 1st and last ratios ,

xdx/y²z = dz/y²

⇒xdx/z = dz

⇒xdx = zdz

Integrating we get,

x²-z² = c2

Hence the required  general solution is  

Φ(x³-y³,x²-y²)where Φ is   arbitrary