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

#### Step-3

Integrating, we have
 $\dfrac{1}{3}\left( {v + \dfrac{4}{3}\ln (3v - 1)} \right) = x + {C_1}$

#### Step-4

Substituting $v=x+2y$ we have
 $x + 2y + \dfrac{4}{3}\ln (3x + 6y - 1) = 3x + {C_2}$ $\Rightarrow y - x + \dfrac{2}{3}\ln (3x + 6y - 1) = C$