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 #