Resolution of Differential equations having the form y' = G(ax + by)

This is another form of differential equation that can be solved using the substitution method. Remember that the substitution method was used in the case of homogeneous differential equations. Homogeneous equations have the form y' = F(y/x).

In order to solve the differential equation  y' = G(ax + by), we use the substitution v = ax + by. Then we have y' = G(v)

Let's derive the equation   v = ax + by

v' = a + by' 

by' = v' - a

y' =  (v' - a)/b

Let's equal both expressions of y'

 (v' - a)/b = G(v)

v' - a = bG(v)

v' = a + bG(v)

dv/dx = a + bG(v)

1/dx = a + bG(v)/dv

dx = [1/a + bG(v)]dv

Let's integrate both sides:

∫dx = 1/a + bG(v∫dv

The resolution of this equation allows to find v. Once we have v we can substitute it in v = ax + by that allows to find y.

Example. Solve the differential equation: y' - (4x - y + 1)² = 0

Let's write v = 4x - y

Then the differential equation becomes: y' - (v + 1)² = 0 (1)

Let's derive v in order to obtain y'. Once we find y' we will have a differential equation in v. We will have to solve this equation for v. Once we find v we can substitute to find y.

v' = 4 - y' then y' = 4 - v'.

Let's substitute y' in (1):

4 - v' - (v + 1)² = 0

4 - (v + 1)² = v'

dv/dx = 4 - (v + 1)²

1/dx = 4 - (v + 1)²/dv

dx = dv/4 - (v + 1)²

Let's integrate both sides:

∫dx = ∫dv/4 - (v + 1)² (2)

Let's calculate the second side. In order to do this, let's transform 1/4 - (v + 1)²:

1/4 - (v + 1)² = 1/-v² -2v + 3

                      = -1/v² + 2v -3

                      = - 1/(v + 3) (v - 1)

   Then the equation (2) becomes:

∫dx = - ∫1/(v + 3) (v - 1) dv (3)

Let's develop 1/(v + 3) (v - 1) in partial fractions:

1/(v + 3) (v - 1) = A/v + 3 + B/v-1

By developing this expression we find A = -1/4 and B = 1/4

By substituting we find :

1/(v + 3) (v - 1) = -1/4( v + 3)  + 1/4(v - 1)

The equation (3) becomes:

 ∫dx  = ∫[ 1/4( v + 3)  - 1/4(v - 1)]dv

 ∫dx = 1/4∫(1/v + 3 -1/v - 1)

x + c = 1/4 ln( v + 3/v - 1) (we don't consider a constant of integration for this integral)

4x + 4c = ln( v + 3/v - 1)

v + 3/v - 1 = e^4x + 4c

                 = e^4x.e^4c

                = (e^4x.)(c) ( we consider 4c to be equivalent to another constant c which we can call also c')

v + 3 = ce^4xv - ce^4x

By doing all the calculations, we find v = c^4x + 3/ce^4x - 1

By substituting v in the expression v = 4x - y we find y = 4x - 3 + ce^4x/-1 + ce^4x

Practice. Solve the differential equation y' = e^9y-x

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