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Find the equation of the curve passing through $(1 , 0 )$ and which has slope $1+\large\frac{y}{x}$ at $(x , y )$

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  • 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$
  • A D.E of first order and first degree is said to be homogeneous if it can be put in the form $\large\frac{dy}{dx}=f\bigg(\large\frac{y}{x}\bigg)$$\;or\; \large\frac{dy}{dx}=\frac{f_1(x,y)}{f_2(x,y)}$ Where $f_1$ and $f_2$ are homogeneous functions in x and y.
  • To solve we put $y=vx$ and proceed.
Step 1:
The slope of the curve $y= f(x)$ at $(x,y)$ is $\large\frac{dy}{dx}$
$\large\frac{dy}{dx}$$=1+ \large\frac{y}{x}$ which is a function of $\large\frac{y}{x}$
Let.$y=vx=>\large\frac{dy}{dx}$$=v+x \large\frac{dv}{dx}$
Step 2:
The D.E becomes
$v+x \large\frac{dv}{dx}$$=1+\large\frac{vx}{x}$
$v+x\large\frac{dv}{dx}$$=1+v$
$dv=\large\frac{dx}{x}$
$\int dv=\int \large\frac{dx}{x}$$+c$
$v=\log x +c$
$\large\frac{y}{x}$$=\log x +c$
The curve passes through $(1,0)$
$\therefore 0= \log 1+c$
=>$c=0$
The equation of the curve is $y=x \log x$
answered Sep 5, 2013 by meena.p
 
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