​$$\text{Which of the following is a solution to }\\ u_x + x^2 u_y = 0 \text{ with } u(x, 0) = e^x?$$​​

Correct option is C

​$$\text{We have, } u_x + x^2 u_y = 0, \; u(x, 0) = e^x. \\[10pt]\text{Using Lagrange's Method:} \\[10pt]\frac{dx}{1} = \frac{dy}{x^2} = \frac{du}{0} \\[10pt]\Rightarrow u = C_1 \\[10pt]\frac{dx}{1} = \frac{dy}{x^2} \quad \Rightarrow \quad x^2 dx = dy \\[10pt]\Rightarrow \frac{x^3}{3} = y + C_2 \\[10pt]\text{So, } \frac{x^3}{3} - y = C_2. \\[10pt]\Rightarrow u = \phi \left( \frac{x^3}{3} - y \right) \text{ is a solution.} \\[10pt]u(x, 0) = e^x. \\[10pt]\text{Using initial conditions: } u(x, 0) = e^x, \text{ let } x = t \text{, then Initial curve : } (t, 0, e^t). \\[10pt]u = C_1 \quad \Rightarrow \quad C_1 = e^t \quad \cdots (1). \\[10pt]\frac{x^3}{3} - y = C_2 \quad \Rightarrow \quad \frac{t^3}{3} = C_2 \quad \Rightarrow \quad t^3 = 3C_2 \quad \Rightarrow \quad t = (3C_2)^{1/3} \quad \cdots (2). \\[10pt]\text{Using (1) and (2), we get:} \\[10pt]C_1 = e^{(3C_2)^{1/3}}. \\[10pt]u = e^{\left( 3 \left( \frac{x^3}{3} - y \right) \right)^{1/3}} \\[10pt]\Rightarrow u(x, y) = e^{(x^3 - 3y)^{1/3}} \quad \text{(Ans.)}.$$​​