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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Solve : $\large \frac{dy}{dx}+\frac{y}{x}=e^x$

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Toolbox:
  • A linear differential equation of the form $\large\frac{dy}{dx}$$=f(x)$ can be solved by seperating the variables and then integrating it.
  • A linear differential equation of the form $\large\frac{dx}{dy}$$+Px=Q$ . has the general solution $xe^{\int pdy}=\int Q e^{\int pdy}.dy+c$
  • $ \int udv=uv-\int vdu$
Step 1:
Given equation is
$\large\frac{dy}{dx}+\frac{y}{x}=e^x$
Clearly this is a linear differential equation of the form
$\large\frac{dy}{dx}$$+Py=Q$
Where $P=\large\frac{1}{x}$ and $Q=e^x$
$\int Pdx=\int \large\frac{1}{x}$$dx=\log x$
$\therefore e^{\large \int Pdx}=e^{\large\log x}=x$
Hence the required solution is
$ye^{\int \large Pdx}=\int Q.e^{\large \int Pdx}dx+c$
Step 2:
$\Rightarrow yx=\int e^x.xdx+c$
Consider $\int e^x.xdx$
This is of the form $\int udv$
Where $u=x$
$du=dx$
$dv=e^xdx$
$v=e^x$
$ \int udv=uv-\int vdu$
$\Rightarrow \int xe^xdx=xe^x-\int e^xdx$
$\qquad\qquad\;\;=xe^x-e^x+c$
Step 3:
$\therefore$ The required solution is
$xy=xe^x-e^x+c$
$xy=e^x(x-1)+c$
answered Oct 3, 2013 by sreemathi.v
 

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