Question

y'=(y(2x+1))/(-x(x+1)) solve show all steps

Answer 1

Do you have any idea how to start?

Answer 2

no

Answer 3

i solved from (2xy+y)dx+(x2+x)dy=0

slove show all steps

Answer 4

and set it equal to y' but now idk what to do

Answer 5

So if you have an equation in the form:
\[y'(t)+p(t)y(t)=g(t) \]
Then the solution is given by:
\[y(t)=c_1 e^{- \int\limits p(t) dt} \int\limits e^{\int\limits p(t) dt} g(t)dt + Ce^{- \int\limits p(t)dt}\]

Answer 6

but i dont have it in that form

Answer 7

You do:
\[y'(t)+\frac{2x+1}{x(x+1)}y(t)=0; p(t)= fraction; g(t)=0\]
Lets call:
\[I =\int\limits \frac{2x+1}{x(x+1)}dx\]
Then the solution is:
\[y(x)=c_1 e^{-I} \int\limits e^I *0 dx+c_2 e^{-I}=c_2 e^{-I}\]
Now lets solve I.
Partial fractions:
\[\frac{2x+1}{x(x+1)}=\frac{A}{x}+\frac{B}{x+1} \implies Ax+A+Bx=2x+1 \implies A=1;B=-1\]
So:
\[I=\int\limits \frac{dx}{x}-\int\limits \frac{dx}{x+1}=\ln|x|-\ln|x+1|\]
So:
\[y(x)=Ce^{-\ln|x|+\ln|x+1|}=Ce^{-\ln|x|}e^{\ln|x+1|}=\frac{C(x+1)}{x}\]

Answer 8

Thats not right...the partial fractions is wrong, B should be +1 So its just:
\[y(x)=C(x(x+1))\]