What is the general solution of the differential equation? $(x+1)y'+y= tan^-1(x) / (x+1)$

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Answer 1 · 2018-01-10T18:53:56

$(1+x)y = int \ tan^-1(x) /(1+x) \ dx + C$

We have:

$(x+1)y'+y= tan^-1(x) / (x+1)$

We can re-arrange this ODE as follows:

$dy/dx + 1/(x+1) y = tan^-1(x)/(1+x)^2$ ..... [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;

$(1+x)dy/dx + y = tan^-1(x) /(1+x)$
$d/dx((1+x)y) = tan^-1(x) /(1+x)$

(Which is incidentally the equation we started with). We can directly integrate to get:

$(1+x)y = int \ tan^-1(x) /(1+x) \ dx + C$

And in terms of elementary functions this is as far we can get:

What is the general solution of the differential equation? (x+1)y'+y= tan^-1(x) / (x+1) (x+1)y'+y= tan^-1(x) / (x+1)