Question

What is the general solution of the differential equation `xdy/dx = y+sinx`?

Answer 1

`y = xCi(x) -sinx + Cx`

Where `Ci(x)` is the Cosine Integral

Explanation:

We have:

`xdy/dx = y+sinx` ..... [A]

This is a First Order Differential Equation which we can manipulate as follows:

`\ \ \ \ \ dy/dx = y/x+sinx/x`

`:. dy/dx - y/x = sinx/x` ..... [B]

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)`

Then the integrating factor is given by;

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

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

`1/xdy/dx - y/x^2 = sinx/x^2`
`:. d/dx(y/x) = sinx/x^2`

Which has now reduced the original differential equation [A] to a First Order separable equation, so we can "seperate the variables" , to get:

`y/x = int \ sinx/x^2 \ dx` ..... [C]

The RHS will prove a challenge, but we can simplify it slightly using a single application of Integration By Parts:

Let `{ (u,=sinx, => (du)/dx,=cosx), ((dv)/dx,=1/x^2, => v,=-1/x ) :}`

Then plugging into the IBP formula:

`int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx`

gives us

`int \ (sinx)(1/x^2) \ dx = (sinx)(-1/x) - int \ (-1/x)(cosx) \ dx`
`:. int \ sinx/x^2 \ dx = -sinx/x + int cosx/x \ dx`

Using this result, we can integrate the earlier result [C] to get:

`y/x = -sinx/x + int cosx/x \ dx + C`

So we have simplified the integral, but we cannot now evaluate this resulting integral any further in terms of elementary functions. Instead we introduce the Cosine Integral:

`Ci(x) := -int_0^oo cost/t \ dt`

Allowing us to write:

`y/x = Ci(x) -sinx/x + C`

`:. y = xCi(x) -sinx + Cx`