Question

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

Answer 1

`y = xe^x + Cx`

Explanation:

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`