## Differential Equations and Linear Algebra (4th Edition)

$y=Cx^2$
Given that: $u=x^2+2y^2$ Differentiate with respect to x: $\frac{du}{dx}=2x+4y\frac{dy}{dx}$ $0=2x+4y\frac{dy}{dx}$ so$\frac{dy}{dx}=-\frac{x}{2y}$ For orthogonal trajectory: $\frac{dy}{dx}=\frac{-1}{f(x,y)}$ $\frac{dy}{dx}=\frac{2y}{x}$ On integrating the above term: $\int\frac{dy}{y}=2\int\frac{dx}{x}$ $\ln{y}=2\ln{x}+c$ $y=e^(2\ln{x})\times e^c$ On simplifying above: $y=x^2\times C$ $y=Cx^2$ Thus, the final answer: $y=Cx^2$