Question

Solve `ydx+x(1-3x^2y^2)dy=0`

Answer 1

`ydx+x(1-3x^2y^2)dy=0`  …………….(1)

Compare the given eqn with Mdx + Ndy=0

∴ M = y  `therefore N=x(1-3x^2y^2)`

`(delM)/(dely)=1` `(delN)/(delx)=1-9x^2y^2`

`(delM)/(dely)!=(delN)/(delx)`

Hence the given diff. eqn is not exact.
But the given diff. eqn is in the form of yf(xy)dx + xf(xy)dy = 0

Integrating factor = `1/(Mx-Ny)=1/(xy-xy+3x^3y^3)=1/(3x^3y^3)`

Multiply the I.F. to eqn (1),

`1/(3x^3y^2)dx+[1/(3x^2y^3)-1/y]dy=0`

`thereforeM_1=1/(3x^3y^2)      N_1=[1/(3x^2y^3)-1/y]`

Now this diff. eqn is exact.
The solution of given diff. eqn is given by,

`intMdx+int[N-del/(dely)Mdx]dy=c`

`intM_1dx=int1/(3x^3y^2)dx=(-1)/(6y^2x^2)`

`del/(dely)intM_1dx=1/(3x^2y^3`

`int[N_1-del/(dely)intM_1dx]dy=int[1/(3x^2y^3)-1/y-1/(3x^2y^3)]dy`

`=int(-1)/ydy=-logy`

`therefore (-1)/(6y^2x^2)-logy=c`