Ordinary differential equations ???

Question

How do you solve this differential equation,

 

 

(x-y² )dx + 2xy dy =0

Answer 1

Let M(x,y) = x - y² and N(x,y) = 2xy. Then ∂M/∂y = -2y and ∂N/∂x = 2y. Since ∂M/∂y ≠ ∂N/∂x, this differential equation is inexact. However, by multiplying the equation by an integrating factor, one may turn the inexact ODE into an exact one. First compute

 

(∂M/∂y - ∂N/∂x)/N = -4y/(2xy) = -2/x.

 

Then, the integrating factor is

 

e^{∫-(2/x) dx} = e^-2ln|x| = 1/x². 

 

Multiplying the differential equation by 1/x² results in 

 

(1)  (1 - y²/x²) dx + 2y/x dy = 0.

 

Since ∂/∂y(1 - y²/x²) = -2y/x² and ∂/∂x(2y/x) = -2y/x², (1) is indeed an exact differential equation. Its solution will be of the form u(x,y) = C, where C is a general constant. So 

 

(2) ∂u/∂x = 1 - y²/x² 

(3) ∂u/∂y = 2y/x.

 

Integrating (1) with respect to x yields 

 

(4) u(x,y) = x + y²/x + f(y),

 

for some function f depending on y. Differentiating (4) with respect to y and comparing with (3), one obtains

 

(5) 2y/x = 1 + 2y/x + f;'(y).

 

From (5), 0 = 1 + f'(y), or -1 = f'(y). Hence f(y) = -y + constant, and the general solution of (1) is

 

x + y²/x - y = C,

 

which is also the general solution of the original differential equation.