Question

Find the solution of the differential equation `x(1+y^2)dx-y(1+x^2)dy =0`?

Answer 1

`y^2 = ( x^2 -1)/2`

Explanation:

We have:

`x(1+y^2)dx-y(1+x^2)dy =0`

If we rearrange this ODE from differential form into standard form we have:

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

`:. y/(1+y^2)dy/dx =x/(1+x^2)`

This is now a separable ODE, do we can "separate the variables" to give:

`int \ y/(1+y^2)dy = int \ x/(1+x^2) \ dx`

We can manipulate the numerators on both sides to get:

`int \ (2y)/(1+y^2)dy = int \ (2x)/(1+x^2) \ dx`

Both integrals are identical and of standard functions, so we can now integrate to get the General Solution:

`ln | 1+y^2 | = ln | 1+x^2 | + C`

We are given that `y=0` when `x=1`, so:

`ln | 1 | = ln | 1+1 | + C +> C = -ln2`

So we have the Particular Solutions

`ln | 1+y^2 | = ln | 1+x^2 | -ln2`

`:. ln ( 1+y^2 ) = ln (( 1+x^2 )/2 )`

`:. 1+y^2 = ( 1+x^2 )/2`

`:. y^2 = ( 1+x^2 )/2 -1`

`:. y^2 = ( x^2 -1)/2`