# Solution by Integrating Factor/Examples/y' + y = x^-1

## Theorem

Consider the linear first order ODE:

$(1): \quad \dfrac {\d y} {\d x} + y = \dfrac 1 x$

with the initial condition $\tuple {1, 0}$.

This has the particular solution:

$y = \ds e^{-x} \int_1^x \dfrac {e^\xi \rd \xi} \xi$

## Proof

This is a linear first order ODE with constant coefficents in the form:

$\dfrac {\d y} {\d x} + a y = \map Q x$

where:

$a = 1$
$\map Q x = \dfrac 1 x$

with the initial condition $y = 0$ when $x = 1$.

 $\ds y$ $=$ $\ds e^{-x} \int_1^x \dfrac {e^\xi \rd \xi} \xi + 0 \cdot e^{x - 1}$ $\ds$ $=$ $\ds e^{-x} \int_1^x \dfrac {e^\xi \rd \xi} \xi$

From Primitive of $\dfrac {e^x} x$ has no Solution in Elementary Functions, further work on this is not trivial.

$\blacksquare$