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Solve the following $ yx^{2}dx+e^{-x}dy=0$

1 Answer

  • 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$
Step 1:
$yx^2dx+e^{-x}dy=0$ divided by $ye^{-x}$
Step 2:
The variable are separated
$ \int e^x x^2 dx +\int \large\frac{dy}{y}$$=c$
Using the Beraoullis formula for the first integral.
$I_1= \int e^xx^2dx=\int u dv $
$u'= 2x$
$u''= 2$
$\quad= uv-u'v_1+u''v_2$
$\quad= x^2e^x-2xe^x +2e^x$
$\quad= e^x[x^2-2x+2]$
The G.S is
$e^x[x^2-2x+2]+\log y=c$
answered Sep 4, 2013 by meena.p