Question

Solve the following differential equation :-
$\\ \dag \boxed{ \sf y \bigg(1 + xy + {x}^{2} {y}^{2} \bigg)dx + x \bigg(1 - xy + {x}^{2} {y}^{2} \bigg)dy = 0} \dag$

Have a great answering time ♡~​

Answer 1

$\tt\sf\rightarrow\: let \: u \: = xy \\ \tt\sf\rightarrow \: then \: y \: = \frac{u}{x} and \: dy \: = (xdu - udx) \div {x}^{2}$

Substituting we get :

$\tt\mapsto \: (x \div {u}^{2} - u + 1)((xdu - udx) \div \: {x}^{2}) + (u \div x) \div {u}^{2} + u + 1)dx = 0 \\ \\ \tt\mapsto \: ((u \div x)( {u}^{2} + u + 1) - (u \div x)( {u}^{2} - u + 1))dx = ( - u {}^{2} + u - du \: 2 {u}^{2} \div x)dx = - {4}^{2} + u - 1)du$

Integrate both sides to obtain : In|x|

$\tt\mapsto \: (- 1 \div 2) \: u + (1 \div 2) \: \: In|x| \: + 1 \div 2u$

Since u = xy

We have In|x|

$\tt\sf\rightarrow - 1 \div 2 \: xy \: \: ( + 1 \div 2)In|x|( + 12xy + c)$

$\\ \dag \boxed{ \sf (-1/2)xy+(1/2)|n|xy|+1/(2xy)+C } \dag$

Answer 2

Answer :-

y = 2y . ln ( xy - x²y² + 1 ) - y. ln ( x )

Refer to the attachment for full explanation ! In the last line I multiplied Wrong of c with y . "c" Doesn't get multiplied and get cancelled !

Note :- I have Learnt Differential Equations Now , So there are chances to mistake . If you found any mistake then kindly report the answer ! ☺️