Question

How do I solve this differential equation? `(x^2 + y^2) dx + xydy = 0`

Answer 1

The GS is:

`y^2 = (A -x^4)/(2x^2)`

Or, alternatively:

`y = +-sqrt(A -x^4)/(sqrt(2)x)`

Explanation:

We have:

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

Which we can write in standard form as:

`dy/dx = -(x^2 + y^2)/(xy)` ..... [1]

Which is a non-separable First Order Ordinary Differential Equation. A suggestive substitution would be to perform a substitution of the form:

`y = xv => dy/dx = v + xv' \ \ \ ` where `v=v(x)`

Then if we substitute into the DE [1], we get

`v + (dv)/dx = -(x^2 + (xv)^2)/(xvx)`

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

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

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

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

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

Which has transformed the initial DE [1] into a separable, DE, so we can Manipulate further, and "separate the variables":

`1/4 \ int \ (4v)/(1+2v^2) \ dv =- \ int \ 1/x \ dx`

And we can now integrate to get:

`1/4 ln |1+2v^2| =- ln|x| + C`

And, noting that `1+2v^2 gt 0 AA x in RR`, and using the properties of logarithms, then

`ln (1+2v^2) =- 4ln|x| + 4C`
`:. ln (1+2v^2) + 4ln|x| - lnA = 0 \ \ \ `, say
`:. ln (1+2v^2) + lnx^4 - lnA = 0`
`:. ln (x^4/A(1+2v^2)) = 0`
`:. x^4/A(1+2v^2) = e^0`

And restoring the substitution, we can now write:

`:. x^4/A(1+2(y/x)^2) = 1`
`:. 1+2(y/x)^2 = A/(x^4)`
`:. 2(y/x)^2 = A/(x^4) -1`
`:. 2y^2/x^2 = (A -x^4)/x^4`

`:. y^2 = (A -x^4)/(2x^2)`

Or, alternatively:

`y = +-sqrt(A -x^4)/(sqrt(2)x)`