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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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The general solution of the differential equation $\large\frac{dy}{dx} $$=e^{\large\frac{x^2}{2}}+xy$ is :\[(A)\;y=ce^{\large\frac{x^2}{2}} \quad (B)\;y=-ce^{\large\frac{x^2}{2}} \quad (C)\;y=(x+c)e^{\large\frac{x^2}{2}} \quad (D)\;y=(c-x)e^{\large\frac{x^2}{2}}\]

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  • If the linear differential equation of the form $\large\frac{dy}{dx}$$+Py=Q$, then the required solution is $ye^{\int pdx}=Q \int e^{\int pdx}.dx+c$
  • Where $e^{\int pdx}$ is the integrating factor (I.F)
Given $\large\frac{dy}{dx}$$=e^{\large\frac{x^2}{2}}$$+xy$
Rearranging this we get,
This is the linear differential equation of the form $\large\frac{dy}{dx}$$+Py=Q$,
Where $P=-x \;and\; Q=e^{\large\frac{x^2}{2}}$
Hence the reqired solution is
$y.e^{\int pdx}=\int Q. e^{\int pdx}+c$
$\int pdx=-\int x dx$
On integrating we get,
Hence $ [e^{\int pdx}]=e^{\Large\frac{-x^2}{2}}$
Hence the required solution is
$y.e^{\Large\frac{-x^2}{2}}=\int e^{\Large\frac{x^2}{2}}.e^{\Large\frac{-x^2}{2}}dx+c$
$y.e^{\large\frac{-x^2}{2}}=\int dx+c$
On integrating we get,
Hence the correct option is $C$
answered May 20, 2013 by meena.p

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