Question

Question #b0376

Answer 1

`(xy)/(x^2-y^2)^2= c_1`

Explanation:

Let `y=vx`, where v is a function of x, which means `dy/dx= v +x (dv)/dx`. The given DE , thus transforms to `v+x(dv)/dx =(v^3 +3v)/(1+3v^2)`

`x(dv)/dx = (v^3+3v)/(1+3v^2) -v =(2v-2v^3)/(1+3v^2)`

`(1+3v^2)/(v(1-v^2))dv= 2(1/x) dx`. Partial fractions on the left side would be

`(1/v +(4v)/(1-v^2))dv= 2 (1/x) dx`. Now integrate on both sides to get

`ln v -2ln(1-v^2) = 2 lnx +C`. On simplification it becomes,

`ln (v/(1-v^2)^2) = ln x^2 +C`
`v/(1-v^2)^2 = c_1x^2`

`(yx^3)/(x^2-y^2)^2= c_1x^2`

`(xy)/(x^2-y^2)^2 =c_1`