solve differential equation using exact method.
(x+y)^2dx+(2xy+x^2-3)dy=0 for the intial condition of y(1)=1

Question

solve differential equation using exact method.
(x+y)^2dx+(2xy+x^2-3)dy=0 for the intial condition of y(1)=1

anonymous · Answer

my answer was xy^2+x^2y-3y+x^3/3

anonymous · Answer

=C

anonymous · Answer

am i right ?

anonymous · Answer

not sure how to apply initial condition

zepdrix · Answer

Ummm yah those are the same terms I'm coming up with!
$$\Large\int\limits (x+y)^2dx+\int\limits (2xy+x^2-3)dy=\int\limits 0$$Integrating each side gives us,$$\Large\sf \frac{1}{3}x^3+xy^2+x^2y-3y=c$$y(1)=1,
you can interpret this as, x=1 when y=1.
So just plug in a bunch of 1's to solve for c,$$\Large\sf \frac{1}{3}+1+1-3=c$$

anonymous · Answer

i dont know what am smoking ....that so passed my mind

anonymous · Answer

dont do DFQ for 2 days n i forget everything

zepdrix · Answer

heh XD

anonymous · Answer

thx u