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

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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:

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


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

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


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

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

Thus from Solution to Linear First Order ODE with Constant Coefficients with Initial Condition:

\(\displaystyle y\) \(=\) \(\displaystyle e^{-x} \int_1^x \dfrac {e^\xi \rd \xi} \xi + 0 \cdot e^{x - 1}\)
\(\displaystyle \) \(=\) \(\displaystyle e^{-x} \int_1^x \dfrac {e^\xi \rd \xi} \xi\)

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


Also see