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## 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.