The solution of the partial differential equation \({{x}^{2}}\frac{\partial z}{\partial x}+{{y}^{2}}\frac{\partial z}{\partial y}=\left( x+y \right)z\) is

Concept:

Xp + Yq = Z

Where \(p\to \frac{dz}{dx}\)

\(q\to \frac{dz}{dy}\) & X, Y and Z are function of x, y & z

Then it’s auxiliary equation will be

\(\frac{dx}{X}=\frac{dy}{Y}=\frac{dz}{Z}\) & on solving above we get u = c₁ & v = c₂

Where c₁ & c₂ = constants u & v are functions of x, y, z

Then solution of the above equation will be

f(u, v) = 0 or u = ϕ (v)

Calculation:

We are given \(\frac{{{x}^{2}}dz}{dx}+\frac{y^2d{{z}}}{dy}=\left( x+y \right)z\)

Then its auxiliary equation will be,

\(\frac{dx}{{{x}^{2}}}=\frac{dy}{{{y}^{2}}}=\frac{dz}{\left( x+y \right)z}\) ---(1)

On solving first two parts of the above equation we

\(\int \frac{dx}{{{x}^{2}}}=\int \frac{dy}{y^2}\Rightarrow -\frac{1}{x}=-\frac{1}{y}+{{c}_{1}}\Rightarrow {{c}_{1}}=\frac{1}{y}-\frac{1}{x}\)

We can write equation (1) as expressed below

\(\frac{\frac{dx}{x}}{x}=\frac{\frac{dy}{y}}{y}=\frac{\frac{dz}{z}}{x+y}\)

\(\Rightarrow \frac{dx}{x}+\frac{dy}{y}=\frac{dz}{z}\)

By integrating the above equation,

In x + In y = In z + In c₂

⇒ In c₂ = In x + In y – In z

\(\Rightarrow {{c}_{2}}=\frac{xy}{z}\)

Hence solution of given equation will be

f(c₁, c₂) = 0