What is the general solution of the differential equation $dy/dx=y+x-1$ ?

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Answer 1 · 2018-03-04T01:46:16

$y= Ce^x - x$

We can use an integrating factor when we have a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

$dy/dx + P(x)y=Q(x)$

We have:

$dy/dx=y+x-1$

$:. dy/dx-y = x-1$ ..... [1]

This is a First Order Ordinary Differential Equation in Standard Form. So we compute and integrating factor, $I$ , using;

$I = e^(int P(x) dx)$
$\ \ = exp(int \ -1 \ dx)$
$\ \ = exp( -x )$
$\ \ = e^( -x )$

And if we multiply the DE [1] by this Integrating Factor, $I$ , we will have a perfect product differential;

$dy/dxe^( -x )-ye^( -x ) = xe^( -x )-e^( -x )$

$:. d/dx{ye^( -x )} = xe^( -x )-e^( -x )$

This has transformed our initial ODE into a Separable ODE, so we can now "separate the variables" to get::

$ye^( -x ) = int \ xe^( -x )-e^( -x ) \ dx$

We can integrate, and we get:

$ye^( -x ) = -xe^-x + C$

Leading to the explicit General Solution:

$y= Ce^x - x$

What is the general solution of the differential equation dy/dx=y+x-1 dy/dx=y+x-1 ?