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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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The general solution of the differential equation$ e^xdy+(ye^x+2x)dx=0\; is$

\[(A)\;xe^y+x^2=\;c\qquad(B)\;xe^y+y^2=\;c\]\[(C)\;ye^x+x^2=\;c\qquad(D)\;ye^y+x^2=\;c\]

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1 Answer

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Toolbox:
  • The solution for first order linear differential equation is $y.(I.F)$=Integration of Q(I.F)+C,where I.F=$e^{\int pdx}$
Step 1:
The given equation can be written as $\large\frac{dy}{dx}$$+y=-2e^{-x}$
Here $P=1$ and $Q=-2xe^{-x}$
Step 2:
Integration of $pdx$=Integration of $dx=x$
Hence $I.F=e^x$
Hence the solution is $ye^x$=Integration of $-2e^{-x}.e^x+C$
$ye^x=-x^2+C$
$ye^x+x^2=C$
Hence $C$ is the correct answer.
answered Jul 29, 2013 by sreemathi.v
 

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