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Show that the equation of the curve whose slope at any point is equal to $y+2x$ and which passes through the origin is $y\;=\;2(e^{x}-x-1)$

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Toolbox:
  • Linear Differential equation.
  • This is of the form $\large\frac{dy}{dx}$$+Py=Q$ where P and Q are functions of x only.
  • The integrating factor $I= e^{\int \large pfd}$ and the $G.S$ is$ ye^{\large pdx}=\int Q e^{\large pdx} dx+c$
Step 1:
Given the slope is $y+2x$
ie $\large\frac{dy}{dx}$$=y+2x$
$\large\frac{dy}{dx}$$-y=2x$
$P(x)=-1,Q(x)=2x$
$e^{\large \int pdx}=e^{\large \int -dx}=e^{-x}$
Step 2:
The solution is $ye^{-x}=\int e^{-x}\;2x\;dx+c$
$ye^{-x}=2[-e^{-x}x -e^{x}]+c$
$ye^{-x}=-2e^{-x}[x+1]+c$
Step 3:
The curve passes through $(0,0)$
$0=-2[0+1]+c$
$c=2$
Step 4:
The equation is
$ye^{-x}=-2e^{-x}(x+1)+c$
$y=-2(x+1)+2e^{-x}$
$y=2[e^{-x}-x-1]$
answered Sep 6, 2013 by meena.p
 
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