Examples On Differential Equations Reducible To Homogeneous Form

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Example – 11

Solve the DE \(\begin{align}\frac{{dy}}{{dx}} = \frac{{2y - x - 4}}{{y - 3x + 3}}.\end{align}\)

Solution: We substitute \(x \to X + h\) and \(y \to Y + k\) where h, k need to be determined :

\[\frac{{dy}}{{dx}} = \frac{{dY}}{{dX}} = \frac{{(2Y - X) + (2k - h - 4)}}{{(Y - 3X) + (k - 3h + 3)}}\]

h and k must be chosen so that

\[\begin{array}{l}2k - h - 4 = 0\\k - 3h + 3 = 0\end{array}\]

This gives \(h = 2\) and \(k = 3.\) Thus,

\[\begin{array}{l}x = X + 2\\y = Y + 3\end{array}\]

Our DE now reduces to

\[\frac{{dY}}{{dX}} = \frac{{2Y - X}}{{Y - 3X}}\]

Using the substitution \(Y = vX,\) and simplifying, we have (verify),

\[\frac{{v - 3}}{{{v^2} - 5v + 1}}dv = \frac{{ - dX}}{X}\]

We now integrate this DE which is VS; the left-hand side can be integrated by the techniques described in the unit on Indefinite Integration.

Finally, we substitute \(v = \frac{Y}{X}\) and

\[\begin{array}{l}X = x - 2\\Y = y - 3\end{array}\]

to obtain the general solution.

Suppose our DE is of the form

\[\frac{{dy}}{{dx}} = f\left( {\frac{{ax + by + c}}{{dx + ey + f}}} \right)\]

We try to find h, k so that

\[\begin{array}{l}ah + bk + c = 0\\dh + ek + f = 0\end{array}\]

What if this system does not yield a solution? Recall that this will happen if  \(\begin{align}\frac{a}{b} = \frac{d}{e}.\end{align}\) How do we reduce the DE to a homogeneous one in such a case ?

Let \(\begin{align}\frac{a}{d} = \frac{b}{e} = \lambda \end{align}\) (say). Thus,

\[\frac{{ax + by + c}}{{dx + ey + f}} = \frac{{\lambda (dx + ey) + c}}{{dx + ey + f}}\]

This suggests the substitution \(dx + ey = v,\) which’ll give

\[\begin{align}&\qquad \;\;d + e\frac{{dy}}{{dx}} = \frac{{dv}}{{dx}}\\&\Rightarrow \quad \frac{{dy}}{{dx}} = \frac{1}{e}\left( {\frac{{dv}}{{dx}} - d} \right)\end{align}\]

Thus, our DE reduces to

\[\begin{align} & \qquad \;\;\frac{1}{e}\left( {\frac{{dv}}{{dx}} - d} \right) = \frac{{\lambda v + c}}{{v + f}}\\&\Rightarrow \quad \frac{{dv}}{{dx}} = \frac{{\lambda ev + ec}}{{v + f}} + d\\ &\qquad \qquad= \frac{{(\lambda e + d)v + (ec + df)}}{{v + f}}\\&\Rightarrow \quad \frac{{(v + f)}}{{(\lambda e + d)v + ec + df}}dv = dx\end{align}\]

which is in VS form and hence can be solved.

Example – 12

Solve the DE \(\begin{align}\frac{{dy}}{{dx}} = \frac{{x + 2y - 1}}{{x + 2y + 1}}.\end{align}\)

Solution: Note that h, k do not exist in this case which can reduce this DE to homogeneous form. Thus, we use the substitution

\[\begin{align}&\qquad \;x + 2y = v\\& \Rightarrow \quad 1 + 2\frac{{dy}}{{dx}} = \frac{{dv}}{{dx}}\end{align}\]

Thus, our DE becomes

\[\begin{align} &\qquad \frac{1}{2}\left( {\frac{{dv}}{{dx}} - 1} \right) = \frac{{v - 1}}{{v + 1}}\\&\Rightarrow \quad \frac{{dv}}{{dx}} = \frac{{2v - 2}}{{v + 1}} + 1\\& \qquad \qquad = \frac{{3v - 1}}{{v + 1}}\\&\Rightarrow \quad \frac{{v + 1}}{{3v - 1}}dv = dx\\&\Rightarrow \quad \frac{1}{3}\left( {1 + \frac{4}{{3v - 1}}} \right)dv = dx\end{align}\]

Integrating, we have

\[\frac{1}{3}\left( {v + \frac{4}{3}\ln (3v - 1)} \right) = x + {C_1}\]

Substituting \(v = x + 2y,\) we have

\[\begin{align}& \qquad \;\; x + 2y + \frac{4}{3}\ln (3x + 6y - 1) = 3x + {C_2}\\&\Rightarrow \quad y - x + \frac{2}{3}\ln (3x + 6y - 1) = C\end{align}\]