Question

What is the solution to the differential equation: `(x^2+y+y^2)dx=xdy` ?

I am trying to separate the variables to LHS and RHS which then I integrate both sides to have `y` in terms of `x`

I can't finish it and was stuck at quite beginning:
`(x^2+y+y^2)/x=dy/dx`

I can't find ways to factorise or group `x` and `y`

Help please, thanks.

Answer 1

`y = xtan(x + c)`

Explanation:

We have the following Differential Equation in differential form

`(x^2+y+y^2) dx = xdy`

Which we can re-arrange as follows:

`dy/dx = (x^2+y+y^2)/x`
`" " = x+y/x+y^2/x`
`" " = x+y/x+y(y/x)`

Try a substitution `v=y/x => y=xv`, then

`dy/dx = v+x(dv)/dx`

Substituting in the DE we get:

`v+x(dv)/dx = x+v+(xv)v`

`:. x(dv)/dx = x+xv^2`

`:. (dv)/dx = 1+v^2`

This is a separable DE, so we can seperate the variables to get:

`int \ 1/(1+v^2) \ dv = int \ dx`

These integrals are standard and so trivial to evaluate:

`arctan v = x + c`
`:. v = tan(x + c)`

And restoring the initial substitution we get:

`y/x = tan(x + c)`

`:. y = xtan(x + c)`

Answer 2

See below.

Explanation:

Making `y = lambda x` and

`dy = lambda dx + x dlambda` we have

`(x^2+y+y^2)dx -x dy -> (x^2+lambda x + x^2 lambda^2) dx - x(x dlambda + lambda dx)`

or

`x((x+lambda+x lambda^2)dx-(x dlambda+lambda dx))=0`

or

`x dx+lambda dx+x lambda^2dx - x dlambda-lambda dx = 0`

or

`x(1+lambda^2)dx-x dlambda = 0`

or

`(dlambda)/(1+lambda^2) = dx`

and after integration

`arctan(lambda) = x + C` or

`lambda = tan(x+C)=y/x` and finally

`y = x tan(x+C)`