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