Question

What is the general solution of the differential equation `(x^2+y^2)dx+(x^2-xy)dy = 0`?

Answer 1

`y/x-2ln(y/x+1) = lnx + C`

Explanation:

We have:

`(x^2+y^2)dx+(x^2-xy)dy = 0`

We can rearrange this Differential Equation as follows:

`dy/dx = - (x^2+y^2)/(x^2-xy)`
`" " = - ((1/x^2)(x^2+y^2))/((1/x^2)(x^2-xy))`
`" " = - (1+(y/x)^2)/(1-y/x)`

So Let us try a substitution, Let:

`v = y/x => y=vx`

Then:

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

And substituting into the above DE, to eliminate `y`:

`v + x(dv)/dx = - (1+v^2)/(1-v)`
`" " = (1+v^2)/(v-1)`

`:. x(dv)/dx = (1+v^2)/(v-1) - v`
`:. " " = {(1+v^2) - v(v-1)}/(v-1)`
`:. " " = {(1+v^2 - v^2+v)}/(v-1)`
`:. " " = (v+1)/(v-1)`

`:. (v-1)/(v+1) \ (dv)/dx = 1/x`

This is now a separable DIfferential Equation, and so "separating the variables" gives us:

`int \ (v-1)/(v+1) \ dv = \ int \ 1/x \ dx`

This is now a trivial integration problem, thus:

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

`v-2ln(v+1) = lnx + C`

And restoring the substitution we get:

`y/x-2ln(y/x+1) = lnx + C`

Answer 2

See below.

Explanation:

Making the substitution

`y = lambda x` we have

`dy = lambda dx + x d lambda` then

`(x^2+y^2)dx+(x^2-xy)dy equiv x^2(1+lambda^2)dx+x^2(1-lambda)(lambda dx + x dlambda)` or assuming `x ne 0`

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

`(1+lambda)dx+x(1-lambda)dlambda=0` This is a separable differential equation so

`((1-lambda)/(1+lambda))dlambda = -dx/x` and integrating both sides

`2log(lambda+1)-lambda = log x + C` or

`(lambda+1)^2/lambda = C_1 x` and solving for `lambda`

`lambda = y/x = 1/2(C_1 x pm sqrt(C_1 x) sqrt(C_1x-4))` then

`y = x /2(C_1 x pm sqrt(C_1 x) sqrt(C_1x-4))`