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

1 Answer

Toolbox:
  • Linear Differential equation.
  • This is of the form $\large\frac{dy}{dx}$$+Py=Q$ where P and Q are functions of x only.
  • The integrating factor $I= e^{\int \large pfd}$ and the $G.S$ is$ ye^{\large pdx}=\int Q e^{\large pdx} dx+c$
Step 1:
$\large\frac{dy}{dx}$$+y=x$
$P(x)=1\quad Q(x)=x$
$e^{\large pdx}=e^{\int \large dx}=e^x$
Step 2:
The GS is $ye^{\large pdx}= \int Q e^{\int \large pdx} dx+c$
$ye^x=\int xe^xdx+c$
$ye^x=\int xe^x-e^x+c$
answered Sep 5, 2013 by meena.p
 
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