ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 8

Tuesday, 5 August 2014

CHAPTER 8 - Examples on Homogeneous DEs

Example: 1

Solve the DE $xdy - ydx = \sqrt {{x^2} + {y^2}} dx$

Solution: 1

Step-1

Upon rearrangement, we have

	$\dfrac{{dy}}{{dx}} = \dfrac{{y + \sqrt {{x^2} + {y^2}} }}{x}$
	$= \left( {\dfrac{y}{x}} \right) + \sqrt {1 + {{\left( {\dfrac{y}{x}} \right)}^2}}$

Step-2

This is obviously a first degree homogeneous DE. We substitute $y=vx$ to obtain:

	$v + x\dfrac{{dv}}{{dx}} = v + \sqrt {1 + {v^2}}$
	$\Rightarrow \dfrac{{dv}}{{\sqrt {1 + {v^2}} }} = \dfrac{{dx}}{x}$

Step-3

Integrating both sides, we have

	$\ln \left\| {v + \sqrt {1 + {v^2}} } \right\| = \ln x + \ln C$
	$= \ln Cx$
	$\Rightarrow \dfrac{y}{x} + \sqrt {1 + \dfrac{{{y^2}}}{{{x^2}}}} = Cx$

Example: 2

Solve the DE $\left( {1 + {e^{x/y}}} \right)dx + {e^{x/y}}\left( {1 - \dfrac{x}{y}} \right)dy = 0$

Solution: 2

Step-1

This DE can be rearranged as

$\dfrac{{dx}}{{dy}} = \dfrac{{{e^{x/y}}\left( {\dfrac{x}{y} - 1} \right)}}{{{e^{x/y}} + 1}}$

Using the substitution $x=vy$ (note : not $y=vx$ ) can reduce this DE to a VS form. (We did not use $y=vx$ since that would’ve led to an expression involving complicated exponentials).

Step-2

We now have

	$v + y\dfrac{{dv}}{{dy}} = \dfrac{{{e^v}\left( {v - 1} \right)}}{{{e^v} + 1}}$
	$\Rightarrow \dfrac{{dy}}{y} = - \dfrac{{{e^v} + 1}}{{{e^v} + v}}dv$

Step-3

Integrating both sides, we have

	$\ln y = - \ln \left\| {{e^v} + v} \right\| + \ln C$
	$\Rightarrow y({e^v} + v) = C$
	$\Rightarrow {e^{x/y}} + \dfrac{x}{y} = \dfrac{C}{y}$

This example should serve to show that $y=vx$ will not always be the most appropriate substitution to solve a homogeneous DE; $x=vy$ could be more appropriate in such a scenario, as in the example above.

Tuesday, 5 August 2014

CHAPTER 8 - Examples on Homogeneous DEs

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