# Differential Equations Reducible To Homogeneous Form

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## Equation reducible to homogeneous form

Many a times, the DE specified may not be homogeneous but some suitable manipulation might reduce it to a homogeneous form. Generally, such equations involve a function of a rational expression whose numerator and denominator are linear functions of the variable, i.e., of the form

$\frac{{dy}}{{dx}} = f\left( {\frac{{ax + by + c}}{{dx + cy + f}}} \right) \qquad \qquad \ldots (1)$

Note that the presence of the constant c and f causes this DE to be non-homogeneous.

To make it homogeneous, we use the substitutions

$\begin{array}{l}x \to X + h\\y \to Y + k\end{array}$

and select h and k so that

$\left. \begin{array}{l}ah + bk + c = 0\\dh + ek + f = 0\end{array} \right\} \qquad \qquad \ldots (2)$

This can always be done $$\left( {{\rm{if }}\frac{a}{b} \ne \frac{d}{e}} \right).$$ The RHS of the DE in (1) now reduces to

\begin{align}&\quad\;\;\; f\left( {\frac{{a(X + h) + b(Y + k) + c}}{{d(X + h) + e(Y + k) + f}}} \right)\\& = f\left( {\frac{{aX + bY + (ah + bk + c)}}{{dX + eY + (dh + ek + f)}}} \right)\\&= f\left( {\frac{{aX + bY}}{{dX + eY}}} \right) \qquad \qquad \qquad (\rm{using\,}(2))\end{align}

This expression is clearly homogeneous! The LHS of (1) is $$\frac{{dy}}{{dx}}$$ which equals \begin{align}\frac{{dy}}{{dY}} \cdot \frac{{dY}}{{dX}} \cdot \frac{{dX}}{{dx}}\end{align}. Since \begin{align}\frac{{dy}}{{dY}} \cdot\frac{{dx}}{{dX}} = 1\end{align}, the LHS $$\frac{{dy}}{{dx}}$$ equals $$\frac{{dY}}{{dX}}.$$ Thus, our equation becomes

$\frac{{dY}}{{dX}} = f\left( {\frac{{aX + bY}}{{dX + eY}}} \right) \qquad\qquad \dots (3)$

We have thus succeeded in transforming the non-homogeneous DE in (1) to the homogeneous DE in (3). This can now be solved as described earlier.

Let us apply this technique in some examples.

Differential Equations
grade 11 | Questions Set 1
Differential Equations
Differential Equations
grade 11 | Questions Set 2
Differential Equations