in differential equations, how did the particular solution of ydx+(2x+3y)dy = 0 became xy^2+y^3=c??? The answer that I got was ln y+1/3ln((x/y)+1)=c....how did that happen?? plssss help

Question

in differential equations, how did the particular solution of ydx+(2x+3y)dy = 0 became xy^2+y^3=c???    The answer that I got was ln y+1/3ln((x/y)+1)=c....how did that happen?? plssss help

anonymous · Answer

will @ganeshie8 know?

anonymous · Answer

the answer I got was ln y + 1/3ln(x/y + 1) = c

ganeshie8 · Answer

did u solve it by substitution ?

anonymous · Answer

yes... I used x=uy; dx=udy+ydu

anonymous · Answer

then I arrived with this answer, lny+1/3ln((x/y) + 1) = c... but the book's answer is xy^2+y^3=c

ganeshie8 · Answer

both answers are equivalent

ganeshie8 · Answer

just simplify the log in ur answer

anonymous · Answer

but how did that happen?? can you show me how??

anonymous · Answer

when I try to simplify, i get $$\ln(y(x/y)+1)^{1/3}=c$$

anonymous · Answer

$$\ln(y(x/y+1)^{1/3})=c$$

anonymous · Answer

how did that turn into xy^2+y^3=c?? @ganeshie8

ganeshie8 · Answer

lny+1/3ln((x/y) + 1) = c

multiply thru by 3 :

3lny + ln(x/y + 1) = C

lny^3 + ln(x/y +1) = C

ln(y^3(x/y)+1) = C

ln(xy^2 + y^3) = C

anonymous · Answer

thnks alot..

ganeshie8 · Answer

yw