Homogeneous Differential Equation

The first question that comes to our mind is what is a homogeneous equation? Well, let us start with the basics. We know that the differential equation of the first order and of the first degree can be expressed in the form Mdx + Ndy = 0, where M and N are both functions of x and y or constants. In particular, if M and N are both homogeneous functions of the same degree in x and y, then the equation is said to be a homogeneous equation. Such an equation can be expressed in the following form:

$$\frac{dy}{dx}$$ = f $$\left(\frac{y}{x}\right)$$

Thus, a differential equation of the first order and of the first degree is homogeneous when the value of $$\frac{dy}{dx}$$ is a function of $$\frac{y}{x}$$. For example, we consider the differential equation:   

                          ($$x^2$$ + $$y^2$$) dy - xy dx = 0

 Now,

                           ($$x^2$$ + $$y^2$$) dy - xy dx = 0 or,   ($$x^2$$ + $$y^2$$) dy - xy dx   

or,  $$\frac{dy}{dx}$$ = $$\frac{xy}{x^2+y^2}$$ = $$\frac{\frac{y}{x}}{1+{\left(\frac{y}{x}\right)}^2}$$ = function of $$\frac{y}{x}$$

Therefore, the equation   ($$x^2$$ + $$y^2$$) dy - xy dx = 0 is a homogeneous equation. On the contrary the differential equation  ($$x^2$$ + $$y^2$$) dy - $$x{y}^2$$  dx = 0 is not a homogeneous equation since in this case, the value of $$\frac{dy}{dx}$$ is not a function of $$\frac{y}{x}$$

Homogeneous Differential Equation Examples

Example 1) Solve ($$x^2$$ - xy) dy = (xy + $$y^2$$)dx 

Solution 1) We have ($$x^2$$ - xy) dy = (xy + $$y^2$$)dx ... (1)

The differential equation (1) is a homogeneous equation in x and y. 

From (1), we have $$\frac{dy}{dx}$$ = $$\frac{xy+y^2}{x^2-xy}$$.... (2)

Now put y = vx, then $$\frac{dy}{dx}$$ = v + x. $$\frac{dy}{dx}$$

From (2), v + x.$$\frac{dy}{dx}$$ = $$\frac{x.vx+v^{2x^2}}{x^2-x.vx}$$ = $$\frac{v+v^2}{1-v}$$

Or,  x $$\frac{dy}{dx}$$ = $$\frac{v+v^2}{1-v}$$ - v = $$\frac{v+v^2-v+v^2}{1-v}$$ = $$\frac{2v^2}{1-v}$$

Or, $$\frac{1-v}{2v^2}$$ dv = $$\frac{dy}{dx}$$ or, $$\frac{dx}{x}$$ = $$\frac{1}{2}$$ $$(\frac{1}{v^2}-\frac{1}{v})$$dv

Integrating, log x = $$\frac{1}{2}$$ $$(-\frac{1}{v}-logv)$$ + $$\frac{1}{2}$$log C

Or, 2 Log x = -  $$\frac{1}{v}$$ - logv + log C or, log $$x^2$$ + log v - log C = - $$\frac{1}{v}$$

OR, Log $$\left(\frac{v{x}^2}{C}\right)$$ = - $$\frac{x}{y}$$  [y = vx] or, $$\frac{v{x}^2}{C}$$ e $$\frac{x}{y}$$, or, xy = Ce - $$\frac{x}{y}$$

Which is the required general solution of homogeneous equation examples?

Example 2) Solve:  ($$x^2$$ + $$y^2$$) dx - 2xy = 0, given that y = 0, when x = 1.

Solution 2)  We have  ($$x^2$$ + $$y^2$$) dx - 2xy dy = 0 or, $$\frac{dy}{dx}$$ = $$\frac{x^2+y^2}{2xy}$$ … (1)

Put y = vx; then $$\frac{dy}{dx}$$ = v + x$$\frac{dv}{dx}$$ 

From, (1), v + x $$\frac{dy}{dx}$$ = $$\frac{x^2+y^2{x}^2}{2x^{2}v}$$ = $$\frac{1+v^2}{2v}$$

Or,   $$\frac{2v}{1-v^2}$$. dv = $$\frac{dx}{x}$$ or, - $$\left(\frac{-2v}{1-v^2}\right)$$dv = $$\frac{dx}{x}$$

Integrating - log(1 - $$v^2$$) = log |x| - log C

Or, Log $$|1-v^2|$$ = - log |x| + log C

Or, log Log $$|1-v^2|$$ x = log C or, ( 1 - $$v^2$$)x = C

Or,  $$(1-\frac{y^2}{x^2})$$ x = C,     or, $$x^2$$ - $$y^2$$ = Cx …(2)

Given y = 0 when x = 1; therefore, from (2), 1 = C

Hence from (2), the required solution is $$x^2$$ - $$y^2$$ = x

Example 3) Solve: $$\{x+ycos\left(\frac{y}{x}\right)\}$$dx = x cos $$\left(\frac{y}{x}\right)$$

Solution 3) $$\{x+ycos\frac{y}{x}\}$$ dx = x cos xcos $$\left(\frac{y}{x}\right)$$dy

OR, $$\frac{dx}{x}$$  = $$\frac{x+ycos\left(\frac{v}{x}\right)}{xcos\left(\frac{v}{x}\right)}$$   put y = vx; then $$\frac{dx}{dx}$$ = v + x $$\frac{dv}{dx}$$.............(1)

From (1), v + x$$\frac{dv}{dx}$$ = $$\frac{x+vxcos\left(\frac{vx}{x}\right)}{xcos\left(\frac{vx}{x}\right)}$$ = $$\frac{1+vcosv}{cosv}$$

Or, v + x $$\frac{dv}{dx}$$ = sec v + v, or, x$$\frac{dv}{dx}$$ = sec v, or, cos v. dv = $$\frac{dx}{x}$$

Integrating both sides, we get $$\int$$ cos v dv = $$\int$$ $$\frac{dx}{x}$$ + C

OR, sin v = log |x| + C, or, sin $$\frac{y}{x}$$ = log|x| + C

Which is the required solution of (1) for the homogeneous equation examples?

Example 4) Find the equation to the curve through (1,0) for which the slope at any point (x, y) is

                                                       $$\frac{(x^2+y^2)}{2xy}$$

Solution 4) for any curve y = f(x), the slope at any point (x,y) is $$\frac{dy}{dx}$$

$$\frac{dy}{dx}$$ = $$\frac{x^2+y^2}{2xy}$$........(1)

Which is a homogeneous differential equation of first order? 

Put y = vx; then $$\frac{dy}{dx}$$ = v + x$$\frac{dv}{dx}$$

From (1), we get v + x  $$\frac{dv}{dx}$$ = $$\frac{x^2+y^2{x}^2}{2x.vx}$$ = $$\frac{1+v^2}{2v}$$  

Or,  x$$\frac{dy}{dx}$$ = $$\frac{1+v^2}{2v}$$ - v = $$\frac{1-v^2}{2v}$$ or, $$\frac{dx}{x}$$   = $$\frac{2v}{1-v^2}$$ dv = $$\frac{-2vdv}{1-v^2}$$

Integrating both sides, we get

log |x| = - $$\int$$  $$\frac{-2vdv}{1-v^2}$$ + lof C = -log $$|1-v^2|$$ + log C

Or,   log |x| = log $$\left|\frac{C}{1-v^2}\right|$$ = log $$\left|\frac{C{x}^2}{x^2-y^2}\right|$$ or, $$\frac{c{x}^2}{x^2-y^2}$$ = x, or, $$x^2$$ - $$y^2$$ = Cx……….(2)

Where C is an arbitrary constant. 

Since this curve passes through the point (1,0);

Therefore, $$1^2$$ - $$0^2$$ = C. 1, or C = 1.

Hence, from (2), the required equation of the curve is $$x^2$$ - $$y^2$$ = x.