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Solve the following $ ydx +xdy=e^{-xy}\;dx $ if it cuts the Y- axis.

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Toolbox:
  • First order , first degree DE
  • Variable separable : Variables of a DE are rearranged to separate then, ie
  • $f_1(x)g_2(y)dx+f_2(x)g_1(y)dy=0$
  • Can be written as $ \large\frac{g_1 (y)}{g_2(y)}$$dy=-\large\frac{f_1(x)}{f_2(x)}$$dx$
  • The solution is therefore $\int \large\frac{g_1(y)}{g_2(y)}$$dy=-\int \large\frac{f_1(x)}{f_2(x)}$$dx+c$
  • Under the variable separable method, identify the following forms: $d(uv)=udv+vdu$
  • and $d\bigg(\large\frac{u}{v}\bigg)=\large\frac{vdu-udv}{v^2}$
$ydx+xdy=e^{-xy}dx$ if it cuts the y-axis
Step:1
$d(xy)=e^{-xy}dx$
Step 2:
$e^{xy}d(xy)=dx$
$\int e^{xy} d(xy)=\int dx+c$
$e^{xy}=x+c$
Step 3:
If t intersects the $y-axis$, x=0
$\therefore c= e^0=1$
The solution is
$e^{xy}=x+1$
answered Sep 4, 2013 by meena.p
 
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