By Florian Cajori

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Hence, the general solution is f x2 − y 2 − u2 , 2xy − u2 = 0, where f is an arbitrary function. 39), we obtain 4C1 = 3C2 . Therefore 4 u2 − x2 + y 2 = 3 2xy − u2 . 37) is given by 7u2 = 6xy + 4 x2 − y 2 . 6. 41) 42 2 First-Order, Quasi-Linear Equations and Method of Characteristics with the Cauchy data u (x, 0) = x2 . The characteristic equations are dy du dx = = . 42) Obviously, dy = −1 dx and du = 1. dx Clearly, x + y = constant = C1 and u − x = constant = C2 . 43) where f is an arbitrary function.

For linear homogeneous ordinary differential equations of order n, a linear combination of n linearly independent solutions is a solution. Unfortunately, this is not true, in general, in the case of partial differential equations. This is due to the fact that the solution space of every homogeneous linear partial differential equation is infinite dimensional. 4) can be transformed into the equation 2uη = 0 by the transformation of variables ξ = x + y, η = x − y. The general solution is u (x, y) = f (x + y) , where f (x + y) is an arbitrary function.

The Cauchy Problem for a Quasi-linear Equation). 12) where 0 ≤ t ≤ 1, and satisfying the condition y0′ (t) a (x0 (t) , y0 (t) , u0 (t)) − x′0 (t) b (x0 (t) , y0 (t) , u0 (t)) = 0. 1) in the neighborhood of C : x = x0 (t), y = y0 (t), and the solution satisfies the initial condition u0 (t) = u (x0 (t) , y0 (t)) , for 0 ≤ t ≤ 1. 13) excludes the possibility that C could be a characteristic. 1. Find the general solution of the first-order linear partial differential equation. 38 2 First-Order, Quasi-Linear Equations and Method of Characteristics x ux + y uy = u.