Question

(y^4+2y)dx + (xy^3 + 2y^4 -4x)dy = 0

Answer 1

Explanation:

by non-exact differential equation we can easily solve this question.

Answer 2

Given,

$(y^{4} +2y)dx + (xy^{3} +2y^{4} -4x)dy= 0$

To Find,

Find the simplify form of this equation

Solution,

$(y^{4} +2y)dx + (xy^{3} +2y^{4} -4x)dy= 0$

$or, (y^{4} +2y)dx= - (xy^{3} +2y^{4} -4x)dy$

$or, \frac{dx}{dy} = \frac{-(xy^{3} +2y^{4} -4x)}{(y^{4}+2y) } \\or, We know that, \frac{u}{v} = \frac{u'v-uv'}{v^{2} }$

$So, \frac{dx}{dy} = (y^{4} +2y)*( -xy^{3} -2y^{4} +4x)' - (-xy^{3} -2y^{4} +4x)*(y^{4} +2y)'/(y^{4} +2y)^{2}$

or, $= \frac{(y^{4}+2y)*(-3y^{2}x+8y^{3} )-(-xy^{2} -2y^{4}+4x)*(4x^{3} +2)}{(y^{4}+2y) ^{2} }$

$= \frac{-3xy^{6}-8y^{7}+6xy^{3}-16y^{4}-4xy^{6} +8xy^{3} +8y^{7} +4y^{4} -16xy^{3} +8 }{y^{8}+4y^{5} + 4y^{2} }$

$= \frac{-7x^{6}-8y^{3} x- 12y^{4}+8 }{y^{8}+4y^{5}+4y^{2} }$

$=\frac{-(7x^{6} + 8xy^{2}+12y^{4}-8) }{(y^{4}+2y)^{2} }$

#SPJ3