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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.
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