# Find the particular solution of the differential equation

• Nov 15th 2010, 01:46 AM
ugkwan
Find the particular solution of the differential equation
Hi again,

$\displaystyle y(2)=2$
$\displaystyle y'+\frac{1}{x}y=0$
$\displaystyle P(x)=\frac{1}{x},Q(x)=0,U(x)=e^{\int\frac{1}{x}dx}$
$\displaystyle y=\frac{1}{U(x)}\int Q(x)U(x)dx$
$\displaystyle y=\frac{1}{x}\int 0 x dx$
$\displaystyle y=\frac{1}{x}\int dx$

Is this the way to go about the problem?
I think this is the set up by first-order linear differential equation. Im having problems with finding the solution.
• Nov 15th 2010, 02:29 AM
Prove It
You have found that your integrating factor is $\displaystyle e^{\int{\frac{1}{x}\,dx}} = e^{\ln{x}} = x$.

So multiplying through by the integrating factor gives

$\displaystyle x\,\frac{dy}{dx} + y = 0$

$\displaystyle \frac{d}{dx}(x\,y) = 0$

$\displaystyle x\,y = \int{0\,dx}$

$\displaystyle x\,y = C$

$\displaystyle y = \frac{C}{x}$.

Now substitute your boundary condition to find $\displaystyle C$.