Question

Solve the equation 7D%29%20dx%20%2B%20%282%20x%5E%7B6%7D-15x%20y%5E%7B6%7D%20%29dy%3D0%20%20" id="TexFormula1" title="(12 x^{5}y+12 x^{6}y-3 y^{5} -18x y^{5}) dx + (2 x^{6}-15x y^{6} )dy=0 " alt="(12 x^{5}y+12 x^{6}y-3 y^{5} -18x y^{5}) dx + (2 x^{6}-15x y^{6} )dy=0 " align="absmiddle" class="latex-formula"> leaving your answer in implicit form.

Answer 1

$\underbrace{(12x^5y+12x^6y-3y^5-18xy^5)}_M\,\mathrm dx+\underbrace{(2x^6-15xy^4)}_N\,\mathrm dy=0$

$M_y=12x^5+12x^6-15y^4-90xy^4$
$N_x=12x^5-15y^4$

$-\dfrac{N_x-M_y}N=6\implies\mu(x)=\exp\left(\displaystyle\int6\,\mathrm dx\right)=e^{6x}$

$\underbrace{(12x^5y+12x^6y-3y^5-18xy^5)e^{6x}}_{\mu M}\,\mathrm dx+\underbrace{(2x^6-15xy^4)e^{6x}}_{\mu N}\,\mathrm dy=0$

You can verify that the partial derivatives are equal.

$F_x=\mu M$
$F=(2x^6y-3xy^5)e^{6x}+f(y)$

$F_y=\mu N$
$(2x^6-15xy^4)e^{6x}+f'(y)=(2x^6-15xy^4)e^{6x}$
$f'(y)=0$
$\implies f(y)=C$

$\implies F(x,y)=(2x^6y-3xy^5)e^{6x}=C$