By Wei-Chau Xie

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**Sample text**

The general solution becomes cos x · sin y = C. General solution Case 2. If sin y = 0, one has y = kπ, k = 0, ±1, ±2, . . It is obvious that y = kπ or sin y = 0 is a solution of the differential equation. However, sin y = 0 is already included in the general solution cos x · sin y = C, with C = 0, obtained in Case 1. 1 the method of separation of variables x y 3 dx + ( y +1)e−x dy = 0. Case 1. If y = 0, separating the variables leads to x ex dx = − y +1 dy. y3 Integrating both sides results in the general solution y +1 dy + C.

Show that equations of the form x dy = f (x y), y dx y = 0, can be converted to variable separable by the transformation x y = v. 2. Using the result obtained above, solve 2+x 2 y 2 x dy . = y dx 2−x 2 y 2 1. Letting x y = v, y+x dy dv = dx dx =⇒ 1 dv x dv x dy = −1= − 1, y dx y dx v dx 30 2 ﬁrst-order and simple higher-order differential equations the differential equation becomes x dv − 1 = f (v) =⇒ v dx x dv = f (v) + 1. v dx Case 1. f (v)+1 = 0. If v0 is a root of f (v0 )+1 = 0, then a solution is x y = v0 .

V = 0 =⇒ dv dv = x − v =⇒ v − 1 = 0. dx dx dv −1 = 0 dx =⇒ dv = dx. Integrating both sides yields v = x + C =⇒ v 2 = (x +C)2 . Replacing v by the original variables results in the general solution x 2 − y 2 = (x +C)2 =⇒ y 2 + 2C x + C 2 = 0. Case 2. v = 0 =⇒ x 2 −y 2 = 0 =⇒ y = ±x. This solution cannot be obtained from the general solution for any value of the constant C and is therefore a singular solution. Combining Cases 1 and 2, the solutions of the differential equation are y 2 + 2C x + C 2 = 0, y = ±x.