Question

How can I solve the differential equation `y'= sinx - xsinx` ?

Answer 1

General solution for your differential equation is:

`y=(x-1)cosx-sinx+C`

Explanation:

.

`y'=dy/dx=sinx-xsinx`

To find `y`, we have to take the integral of `y'`:

`y=int(sinx-xsinx)dx`

`y=intsinxdx-intxsinxdx=-cosx-I`

`I=intxsinxdx`

The argument of the integral is product of two functions. As such, we will use integration by parts:

`u=x, and dv=sinxdx`

`du=dx, and v=-cosx`

`intudv=uv-intvdu`

`intxsinxdx=-xcosx-int-cosxdx=`

`-xcosx+intcosxdx=-xcosx+sinx`

`I=-xcosx+sinx`

Let's plug it in:

`y=-cosx-(-xcosx+sinx)=-cosx+xcosx-sinx`

`y=(x-1)cosx-sinx+C`

This is the solution to the differential equation in your problem statement.

Answer 2

The General Solution is:

`y = xcosx - sinx -cosx + C`

Explanation:

We have:

`y' = sinx-xsinx`

Using Leibniz's Notation we can write this as:

`dy/dx = sinx-xsinx`

Which is a First Order Ordinary Differential Equation, so we can "separate the variables" to get:

`int \ dy = int \ sinx-xsinx \ dx`

Leading to:

`y = int \ sinx \ dx - int \ xsinx \ dx + C`

The first integral is trivial, and we require Integration By Parts for the Second integral, as follows:

Let # { (u,=x, => (du)/dx,=1), ((dv)/dx,=sin x, => v,=-cosz ) :}#

Then plugging into the IBP formula:

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

We have:

`int \ (x)(sinx) \ dx = (x)(-cosx) - int \ (-cosx)(1) \ dx`

`:. int \ xsinx \ dx = -xcosx + sinx`

Using this result we return to the DE solution to get:

`y = -cosx - {-xcosx + sinx}+ C`
`\ \ = -cosx + xcosx - sinx+ C`
`\ \ = xcosx - sinx -cosx + C`

Which is the General Solution.