The differential equation $y-3y^4=(y^3+3x)y\text{'}|$ can $be$ written $in$ differential form: $M(x,y)dx+N(x,y)dy=0|$ where $M(x,y)$ and $N(x,y)=$ The term $M(x,y)dx+N(x,y)dy|$ becomes $an$ exact differential $if$ the left hand side above $is$ divided $by{y}^4|.$ Integrating that new equation, the solution $of$ the differential equation $is$

$(1$ point$)$ The differential equation

$y+y^4=(y^3+3x)y^{\text{'}}$

can $be$ written $in$ differential form:

$M(x,y)dx+N(x,y)dy=0$

where

$M(x,y)=$

$,$ and $N(x,y)=M(x,y)dx+N(x,y)dy$ becomes $an$ exact differential $if$ the left hand side above

$is$ divided $by{y}^4.$ Integrating that new equation, the solution $of$ the differential equation $is$

$=C.$