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

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Answer 1 · 2018-01-06T16:14:52

# y = (1+x)ln(1+x) + C(1+x) #

We have:

# (1+x)dy/dx-y=1+x #

We can re-arrange this ODE as follows:

# dy/dx - 1/(1+x) y=1 # ..... [1]

This is a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

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

We can readily generate an integrating factor when we have an equation of this form, given by;

# I = e^(int P(x) dx) #
# \ \ = exp(int \ -1/(1+x) \ dx) #
# \ \ = exp( -ln(1+x) ) #
# \ \ = -(1+x) #

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

# -1/(1+x) dy/dx + 1/(1+x)^2 y = -1/(1+x) #
# :. d/dx[ -1/(1+x) y] = -1/(1+x) #

Which we can directly integrate to get:

# 1/(1+x) y = int \ 1/(1+x) \ dx #
# :. 1/(1+x) y = ln(1+x) + C #

# :. y = (1+x)ln(1+x) + C(1+x) #