What is the general solution of the differential equation $y'=y/x+xe^x$ ?

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Answer 1 · 2017-12-23T21:15:23

$y = xe^x + Cx$

We seek a solution to the First Order ODE:

$y'=y/x+xe^x$

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

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

So rewrite the equations in standard form as:

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

Then the integrating factor is given by;

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

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

$1/xdy/dx - y/x^2 = e^x$

$:. d/dx( y/x ) = e^x$

Which we can directly integrate to get:

$y/x = int \ e^x + C$

$:. y/x = e^x + C$

$:. y = xe^x + Cx$

What is the general solution of the differential equation y'=y/x+xe^x y'=y/x+xe^x ?