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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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The solution of $\Large \frac{dy}{dx}\normalsize +y=e^{-x},y(0)=0$ is:\begin{array}{1 1}(A)\;y=e^x(x-1) & (B)\;y=xe^{-x}\\(C)\;y=xe^{-x}+1 & (D)\;y=(x+1)e^{-x}\end{array}

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  • A linear differential equation of the form $\large\frac{dy}{dx}$$+Py=Q$ has the general solution as $ye^{\int pdx}=\int Q.e^{\int pdx}.dx+c$
Given $ \large\frac{dy}{dx}$$+y=e^{-x}$
This is a linear differential equation of the form $\large\frac{dy}{dx}+Py=Q$, where $P=1$ and $Q=e^{-x}$
Hence the required solution is $ye^{\int pdx}=\int Q e^{\int pdx}$$+c$
Where $e^{\large\int pdx}$ is the integrating factor (I.F)
$\int pdx=\int 1.dx$
Hence $I.F\;e^{\large\int pdx}=e^x$
The required solution is
$ye^x=\int e^{-x}.e^xdx+c$
$=\int dx+c$
It is given $y(0)=0$ which implies
When $x=0,y=0$
Substituting for x and y
Therefore the required solution is
$ye^{x}=x$ or $ y=xe^{-x}$
Hence $B$ is the correct option
answered May 16, 2013 by meena.p

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