dy/dx =[ (e^y)(sin x)^2 ] / [y (sec x)] sorry about the mess, i'm still learning how to use latex
Follow Math Help Forum on Facebook and Google+
Originally Posted by crafty dy/dx =[ (e^y)(sin x)^2 ] / [y (sec x)] sorry about the mess, i'm still learning how to use latex Your equation separates so Just need to carry out the integrations.
(e^(-y))(-y - 1) = (1/3)(sin x)^3 + C this is what i get but do i have to put this in terms of y and if so how would i do that?
View Tag Cloud