ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 10

Tuesday, 5 August 2014

CHAPTER 10 - Worked Out Examples

Example: 1

Solve the DE $\dfrac{{dy}}{{dx}} = \dfrac{{2y - x - 4}}{{y - 3x + 3}}$

Solution: 1

Step-1

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

$\dfrac{{dy}}{{dx}} = \dfrac{{dY}}{{dX}} = \dfrac{{(2Y - X) + (2k - h - 4)}}{{(Y - 3X) + (k - 3h + 3)}}$

$h$ and $k$ must be chosen so that

	$2k - h - 4 = 0$
	$k - 3h + 3 = 0$

This gives $h=2$ and $k=3$ . Thus,

	$x = X + 2$
	$y = Y + 3$

Step-2

Our DE now reduces to

$\dfrac{{dY}}{{dX}} = \dfrac{{2Y - X}}{{Y - 3X}}$

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

$\dfrac{{v - 3}}{{{v^2} - 5v + 1}}dv = \dfrac{{ - dX}}{X}$

Step-3

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 $v = \dfrac{Y}{X}$ and

	$X = x - 2$
	$Y = y - 3$

to obtain the general solution.

Suppose our DE is of the form

$\dfrac{{dy}}{{dx}} = f\left( {\dfrac{{ax + by + c}}{{dx + ey + f}}} \right)$

We try to find $h,k$ so that

	$ah + bk + c = 0$
	$dh + ek + f = 0$

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

Let $\dfrac{a}{d} = \dfrac{b}{e} = \lambda$ (say). Thus,

$\dfrac{{ax + by + c}}{{dx + ey + f}} = \dfrac{{\lambda (dx + ey) + c}}{{dx + ey + f}}$

This suggests the substitution $dx + ey = v$ , which’ll give

	$d + e\dfrac{{dy}}{{dx}} = \dfrac{{dv}}{{dx}}$
	$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{e}\left( {\dfrac{{dv}}{{dx}} - d} \right)$

Thus, our DE reduces to

	$\dfrac{1}{e}\left( {\dfrac{{dv}}{{dx}} - d} \right) = \dfrac{{\lambda v + c}}{{v + f}}$
	$\Rightarrow \dfrac{{dv}}{{dx}} = \dfrac{{\lambda ev + ec}}{{v + f}} + d$
	$= \dfrac{{(\lambda e + d)v + (ec + df)}}{{v + f}}$
	$\Rightarrow \dfrac{{(v + f)}}{{(\lambda e + d)v + ec + df}}dv = dx$

which is in VS form and hence can be solved.

Example: 2

Solve the DE $\dfrac{{dy}}{{dx}} = \dfrac{{x + 2y - 1}}{{x + 2y + 1}}$

Solution: 2

Step-1

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

	$x + 2y = v$
	$\Rightarrow 1 + 2\dfrac{{dy}}{{dx}} = \dfrac{{dv}}{{dx}}$

Step-2

Thus, our DE becomes

	$\dfrac{1}{2}\left( {\dfrac{{dv}}{{dx}} - 1} \right) = \dfrac{{v - 1}}{{v + 1}}$
	$\Rightarrow \dfrac{{dv}}{{dx}} = \dfrac{{2v - 2}}{{v + 1}} + 1$
	$= \dfrac{{3v - 1}}{{v + 1}}$
	$\Rightarrow \dfrac{{v + 1}}{{3v - 1}}dv = dx$
	$\Rightarrow \dfrac{1}{3}\left( {1 + \dfrac{4}{{3v - 1}}} \right)dv = dx$