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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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The solution of $x \large\frac{dy}{dx}$$+y=e^x$ is:\[(A)\;y=\frac{e^x}{x}+\frac{k}{x} \quad (B)\;y=xe^x+cx \quad (C)\;y=xe^x+k \quad (D)\;x=\frac{e^y}{y}-\frac{k}{y}\]

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  • A linear differential equation of the form $\large \frac{dy}{dx}$$+Py=Q$ has a general solution $y e ^{\int pdx}=\int Q. e^{\int pdx}.dx+c$
  • where $e^{\int pdx}$ is the integrating factor (I.F)
  • $ e^{\large \log x}=x$
Given $ x \large\frac{dy}{dx}$$+y=e^x$
Divide throughout by $x$ we get
Clearly this is a linear differential equation of the form $\large\frac{dy}{dx}$$+Py=Q,$, where $P=\large\frac{1}{x}$ and $Q=\large\frac{e^x}{x}$
Hence the required solution is
$y e ^{\int pdx}=\int Q. e^{\int pdx}+c$
Let us now find $e^{\int pdx}$ which is the integrating factor (I.F)
$\int \large\frac{1}{x} $$dx=\log x$
Hence $e^{ log x}=x$ Therefore $I.F=x$
Hence the required solution is
$y.x=\int \large\frac{e^x}{x}$$.x+k$
$=>xy=\int e^x+c$
On integrating we get,
Hence the correct option is $A$
answered May 17, 2013 by meena.p

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