Question

Solve the initial-value problem y'=(xy+y^2)/x^2, y(-1)=2.

Answer 1

@SithsAndGiggles

Answer 2

y=ux
u=y/x
y'=u+u'x
y'=u+u^2
u'x=u^2
du/u^2=dx/x
-1/u+ln abs(x)+C
u=-1/(ln abs(x)+C)
y=-x/(ln abs(x)+C)
y(-1)=1/(0+C)=2
C=1/2
y=-2x/(ln abs(x)+1)

Answer 3

Which phrase represents the algebraic expression 5x - 9