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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Solve the following differential equation : $ xdy-(y-x^3)dx=0 $

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Toolbox:
  • If the equation is of the form $\large\frac{dy}{dx}$$ + py= Q$, then the integrating factor is $e^{\large\int pdx}$
Step 1:
$xdy-(y-x^3)dx=0$
This can be written as
$xdy=(y-x^3)dx$
divide throughout by $x$
$\large\frac{dy}{dx}=\frac{y}{x}$$-x^2$
$\large\frac{dy}{dx}-\frac{y}{x}$$=-x^2$
This is a linear differential equation of the form
$\large\frac{dy}{dx}$$+Py=Q$
Step 2:
The integrating factor I.F is
$e^{\int Pdx}=e^{-\large\frac{1}{x}dx}$
$\qquad=e^{\large -\log x}$
$\qquad=e^{\large \log (1/x)}$
$\qquad=\large\frac{1}{x}$
Step 3:
The required solution is
$ye^{\int Pdx}=\int Qe^{\int Pdx}.dx+c$
$y.\big(\large\frac{1}{x})=$$-\int x^2\times \large\frac{1}{x}$$dx+c$
$\large\frac{y}{x}$$=-\int x\times dx+c$
$\large\frac{y}{x}=\frac{-x^2}{2}$$+c$
$\large\frac{y}{x}+\frac{x^2}{2}$$=c$
$2y+x^3=2cx$
(i.e) $x^3-2cx+2y=0$ is the required solution.
answered Nov 14, 2013 by sreemathi.v
 
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