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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
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