utnip123
- 9
- 0
xy'=yln(xy)
xdy=yln(xy)dx
\frac{dy}{y}=\frac{ln(xy)dx}{x}
Substitution:
v=ln(xy)
dv = \frac{dy}{y}-\frac{dx}{x}
dv-\frac{dx}{x}=\frac{vdx}{x}
∫\frac{dv}{v+1}=∫\frac{dx}{x}
ln(v+1)=ln x + C
v+1 = Cx
ln(xy) +1 = Cx
Would that basically be the complete answer?
xdy=yln(xy)dx
\frac{dy}{y}=\frac{ln(xy)dx}{x}
Substitution:
v=ln(xy)
dv = \frac{dy}{y}-\frac{dx}{x}
dv-\frac{dx}{x}=\frac{vdx}{x}
∫\frac{dv}{v+1}=∫\frac{dx}{x}
ln(v+1)=ln x + C
v+1 = Cx
ln(xy) +1 = Cx
Would that basically be the complete answer?