# Differential Equations Reducible to Homogeneous Form

## 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.