Homogeneous Differential Equations

 An homogeneous differential equation is one that can be written as y' = F(y/x). The equation is not written in this form most of the times. You have to transform it in order to write it under this form. This type of equation is solved by substitution. That's another type of equation that's solved by substitution. Remember that in the case of a Bernoulli differential equation we use the substitution v = y¹⁻ⁿ. In the case of an homogenous differential equation we are going to see what type of substitution we can use.

Let's bring this homogeneous differential equation to us in order to solve it. Here it comes as y' = F(y/x). It seems natural that we can substitute y/x by something. Let's call it v. We can write v =  y/x. Then we have: y = vx.

Let's derive both sides of the equation:

y' = v'x + v

Remember also that the differential equation is y' = F(y/x) or y' = F(v)

 Equaling both expressions of y' we have:

v'x + v = F(v)

We have a differential equation with a new dependent variable v. This equation can be solved by the method of separation of variables. Let's continue;

v'x = F(v) - v

(dv/dx)x = F(v) - v

x/dx = F(v) - v/dv

dx/x  = dv/F(v) - v

Now that the variables are separated we can integrate both sides of the equation in order to find v. Once we find v we can use the substitution to find y.

Example. Solve the differential equation xyy' + 4x² + y² = 0

Let's bring the expression y/x in the equation by dividing by x. Let's divide both sides by x²:

 xyy'/x² + 4x²/x² +  y²/x²  = 0

yy'/x + 4 + (y/x)² = 0

y/xy' + 4 + (y/x)² = 0

Let's substitute y/x by v:

vy' + 4 + v² = 0  (1)

Let's find y' in order to have an equation with the new variable v:

y/x = v 

y = vx

y' = v'x + v

Let's substitute y' in equation (1)

v(v'x + v) + 4 + v² = 0

vv'x + v² + 4 +  v² = 0

vv'x + 2v² + 4 = 0

vv'x  = -2v² - 4 

Let's divide both sides by v in order to separate the variables:

v'x  = -2v² - 4/v

(dv/dx)x = -2v² - 4/v

x/dx = -2v² - 4/vdv

dx/x = -vdv/ 2v² + 4

Now that the variable are separated let's integrate both sides in order to find v;

∫dx/x = ∫(-v/ 2v² + 4)dv = -∫(v/ 2v² + 4)dv (2)

Let's solve the integral in the second side by doing the substitution u =  2v² +4

du = 4vdv dv = du/4v

The integral at the second side becomes:

∫(v/ 2v² +4)dv = ∫(v/u)(du/4v) = ∫du/4u = 1/4∫du/u = 1/4lnu = 1/4ln (2v² +4) + k

Let's substitute the value of the integral we just calculated in (2)

∫dx/x = -1/4ln (2v² + 4) + k

lnx + K' = -1/4ln (2v² +4) + k

1/4ln (2v² + 4) = -lnx - k' + k

Let's call c the expression-k'+k 

1/4ln (2v² + 4) = -lnx + c

1/4ln (2v² + 4) = -lnx + c

ln(2v² + 4)^1/4  =  -lnx + c

(2v² + 4)^1/4 = e^(-lnx + c)

                     = (e^-lnx)(e^c)

                     = (e^lnx^-1)(d) We call e^c d

  e^lnx^-1 being equal to x^-1, we have:

(2v² + 4)^1/4 = (x^-1)(d) 

Let's raise both sides to the power of 4:

2v² + 4 = x⁻⁴d⁴

Let's call d⁴ g:

2v² + 4 = x⁻⁴g

            = g/x⁴

v²  = g-4x⁴/2x⁴

v = ∓(⎷2/2) (⺁g-4x⁴)/x²

By substituting v in the expression v = y/x, we obtain:

y = ∓⎷2/2⺁g-4x⁴/x

Practice. Solve the differential equation for the initial condition y(2) = -7 and x>0