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Solve the following. $\large\frac{dy}{dx}$$+xy\;=\;x$

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Toolbox:
  • Linear Differential equation.
  • This is of the form $\large\frac{dy}{dx}$$+Py=Q$ where P and Q are functions of x only.
  • The integrating factor $I= e^{\int \large pfd}$ and the $G.S$ is$ ye^{\large pdx}=\int Q e^{\large pdx} dx+c$
Step 1:
$\large\frac{dy}{dx}$$+xy=x$
$P(x)=x\;Q(x)=x$
$e^{ \large \int pdx}=e^{\large\int xdx}$
$\qquad=e^{\large\frac{x^2}{2}}$
The solution is $ye^{\large\frac{x^2}{2}}=\int xe^{\Large\frac{x^2}{2}}dx+c$
$ye^{\large\frac{x^2}{2}}=e^{\large\frac{x^2}{2}}$$+c$
answered Sep 6, 2013 by meena.p
 
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