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Tuesday, 5 August 2014

CHAPTER 14 - Examples on Exact DEs

   Example: 1     

Solve the DE \dfrac{{xdy}}{{{x^2} + {y^2}}} = \left( {\dfrac{y}{{{x^2} + {y^2}}} - 1} \right)dx
Solution: 1

Step-1

Upon rearrangement, this DE gives
\dfrac{{xdy - ydx}}{{{x^2} + {y^2}}} =  - dx\ldots (1)

Step-2

From the last page, the L.H.S of (1) is the exact differential d\left( {{{\tan }^{ - 1}}\dfrac{y}{x}} \right). Thus, our DE reduces to
d\left( {{{\tan }^{ - 1}}\dfrac{y}{x}} \right) + dx = 0
Integrating, we obtain the solution as
{\tan ^{ - 1}}\dfrac{y}{x} + x = C

However, it is very likely that we won’t be able to make out just be inspection whether the DE is exact or not. If the DE is not exact, it can be rendered exact by multiplying it with an integrating factor I.F. In the case of the first-order linear DE
\dfrac{{dy}}{{dx}} + Py = Q
the I.F. {e^{\int {Pdx} }} renders the DE exact:
\dfrac{d}{{dx}}\left( {y{e^{\int {Pdx} }}} \right) = Q{e^{\int {Pdx} }}
and the solution is now obtainable by integration.
If fact, a systematic approach exists to determine the I.F. in a general case (if such an I.F. is possible at all.). However, we’ll not be discussing that approach here since it is beyond our current scope. Let us see another example, where the solution is easily obtained by the recognition of exact differentials present in the equation.
     Example: 2    

Solve the DE x\cos \left( {\dfrac{y}{x}} \right)\left( {ydx + xdy} \right) = y\sin \left( {\dfrac{y}{x}} \right)\left( {xdy - ydx} \right)
Solution: 2

Step-1

Upon rearrangement, we have
ydx + xdy = \dfrac{y}{x}\tan \dfrac{y}{x}\left( {xdy - ydx} \right)
 = xy\tan \dfrac{y}{x}\left( {\dfrac{{xdy - ydx}}{{{x^2}}}} \right)
 \Rightarrow \dfrac{{ydx + xdy}}{{xy}} = \tan \dfrac{y}{x}\left( {\dfrac{{xdy - ydx}}{{{x^2}}}} \right)

Step-2

Using the results on the last page, this can be written as
\dfrac{{d(xy)}}{{xy}} = \tan \left( {\dfrac{y}{x}} \right)d\left( {\dfrac{y}{x}} \right)

Step-3

The solution is now obtained simply by integrating both sides :
\ln (xy) = \ln \left( {\sec \left( {\dfrac{y}{x}} \right)} \right) + \ln C
 \Rightarrow  xy = C\sec \left( {\dfrac{y}{x}} \right)

As described in the introduction, differential equations are so important for the very reason that they find a wide application in studying all sorts of scientific phenomena. The motion of a body in a force field, radioactive decay and population growth were some of the phenomena mentioned that must use DEs for analysis. In some of the subsequent solved examples, we apply the DE- solving techniques that we’ve learnt in the previous section, to solve practical and interesting problems.
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