**Differential Equations Reducible to Homogeneous Form**

A differential equation of the form \(\frac{dy}{dx}=\frac{ax+by+c}{{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}}\), where \(\frac{a}{{{a}_{1}}}\ne \frac{b}{{{b}_{1}}}\) can be reduced to homogeneous form by taking new variable x and y such that x = X + h and y = Y + k, where h and k are constants to be so chosen as to make the given equation homogeneous. With the above substitutions, we get dx = dX and dy = dY, so that \(\frac{dy}{dx}=\frac{dY}{dX}\).

Hence, the given equation becomes

\(\frac{dY}{dX}=\frac{a\left( X+h \right)+b\left( Y+k \right)+C}{{{a}_{1}}\left( X+h \right)+{{b}_{1}}\left( Y+k \right)+{{C}_{1}}}\).

\(=\frac{aX+bY+\left( ah+bk+C \right)}{{{a}_{1}}X+{{b}_{1}}Y+\left( {{a}_{1}}h+{{b}_{1}}k+{{C}_{1}} \right)}\).

Now, choose h and k such that

Ah + bk + C = 0 and a₁h + b₁k + C₁ = 0

Then, the differential equation becomes

\(\frac{dY}{dX}=\frac{aX+bY}{{{a}_{1}}X+{{b}_{1}}Y}\), which is homogeneous.

Now, this equation can be solved as in homogeneous equations by substituting Y = υX. Finally, by replacing X by (x – h) and Y by (x – k) we shall get the solution in original variables x and y.

If however \(\frac{a}{{{a}_{1}}}=\frac{b}{{{b}_{1}}}=m\) (say), then the differential equation becomes of the form \(\frac{dy}{dx}=\frac{m\left( {{a}_{1}}x+{{b}_{1}}y \right)+C}{{{a}_{1}}x+{{b}_{1}}y+{{C}_{1}}}\).

To solve such a differential equation put υ = a₁x + b₁y, get rid of y and then the transformed equation will be such that the variables are separable.

**Example:** Find the solution of the differential equation \(\frac{dy}{dx}=\frac{x+y+3}{2x+2y+1}\).

**Solution: **Here, \(\frac{a}{{{a}_{1}}}=\frac{b}{{{b}_{1}}}=\frac{1}{2}\) i.e., the coefficients of x and y in the Nr and Dr of the expression for \(\frac{dy}{dx}\) are proportional. Proper substitution in this case, therefore, will be to put v for x + y. Let x + y = v. Then, \(1+\frac{dy}{dx}=\frac{dv}{dx}\) with these substitutions the given equation reduces to \(\frac{dv}{dx}-1=\frac{v+3}{2v+1}\) or \(\frac{dv}{dx}=\frac{v+3}{2v+1}+1=\frac{3v+4}{2v+1}\).

Or \(dx=\frac{2v+1}{3v+4}dv=\left[ \frac{2}{3}-\frac{\frac{5}{3}}{3v+4} \right]dv\).

∴ On integrating, \(x+C=\frac{2}{3}v-\frac{5}{3}.\frac{1}{3}\log \left( 3v+4 \right)\).

\(\Rightarrow x+C=\frac{2}{3}v-\frac{5}{9}\log \left( 3v+4 \right)\).

Or \(x+C=\frac{2}{3}\left( x+y \right)-\left( \frac{5}{9} \right)\log \left( 3x+3y+4 \right)\), (∵ v = x + y)

Which is the required solution.