Question

The complete solution of \[ y^3\frac{\partial u}{\partial x} + x^2\frac{\partial u}{\partial y} = 0 \] is \(u(x,y)=\)

• A. \(Ae^{K(x^3-y^4)}\)
• B. \(Ae^{K(x^2+y^3)}\)
• C. \(Ae^{K\frac{x^3}{y^4}}\)
• D. \(Ae^{K\left(\frac{x^3}{3}-\frac{y^4}{4}\right)}\)

Hint:

For a Lagrange PDE, \[ P\frac{\partial u}{\partial x} + Q\frac{\partial u}{\partial y} = R, \] solve the auxiliary equations \[ \boxed{ \frac{dx}{P} = \frac{dy}{Q} = \frac{du}{R}. } \] The complete solution is obtained from the characteristic integrals.

Answer 1

The Correct Option is D

Step 1: Form the auxiliary equations. For the Lagrange's linear partial differential equation \[ P\frac{\partial u}{\partial x} + Q\frac{\partial u}{\partial y} = 0, \] the auxiliary equations are \[ \frac{dx}{P} = \frac{dy}{Q} = \frac{du}{0}. \] Here, \[ P=y^3, \qquad Q=x^2. \] Hence, \[ \frac{dx}{y^3} = \frac{dy}{x^2} = \frac{du}{0}. \]

Step 2: Obtain the first integral. Using \[ \frac{dx}{y^3} = \frac{dy}{x^2}, \] we get \[ x^2\,dx = y^3\,dy. \] Integrating, \[ \frac{x^3}{3} = \frac{y^4}{4} + C. \] Therefore, \[ \frac{x^3}{3} - \frac{y^4}{4} = C. \]

Step 3: Write the complete solution. Since \[ \frac{du}{0}, \] \(u\) remains constant along the characteristic curves. Hence, \[ u = \phi\!\left( \frac{x^3}{3} - \frac{y^4}{4} \right). \] Among the given options, this is represented by \[ u = Ae^{K\left(\frac{x^3}{3}-\frac{y^4}{4}\right)}. \] Therefore, \[ \boxed{ u = Ae^{K\left(\frac{x^3}{3}-\frac{y^4}{4}\right)} } \] is the correct answer. Thus, \[ \boxed{(D)} \] is the correct answer.