- #1
proto
- 5
- 0
[tex]\frac{xy'}{(\ln x\arctan y)-1}=(1+y^2)\arctan y\\[/tex]
[tex]t=\arctan y\\[/tex]
[tex]t'=\frac{1}{1+y^2}y'\\[/tex]
[tex]\frac{x}{\ln (x)t-1}=\frac{t}{t'}\\[/tex]
[tex]\frac{x}{\ln (x)t-1}=\frac{tdx}{dt}\\[/tex]
[tex]xdt=(\ln (x)t-1)tdx\\[/tex]
[tex]\frac{dt}{\ln (x)t-1}=\frac{dx}{x}\\[/tex]
still can't beak it as one type of variable on each side
so i substitute by another variable
[tex]z=\ln x\\[/tex]
[tex]dz=\frac{dx}{x}\\[/tex]
[tex]\frac{dt}{zt-1}=dz\\[/tex]
i don't know how to separate each variable type on one side
so i could integrate
??
[tex]t=\arctan y\\[/tex]
[tex]t'=\frac{1}{1+y^2}y'\\[/tex]
[tex]\frac{x}{\ln (x)t-1}=\frac{t}{t'}\\[/tex]
[tex]\frac{x}{\ln (x)t-1}=\frac{tdx}{dt}\\[/tex]
[tex]xdt=(\ln (x)t-1)tdx\\[/tex]
[tex]\frac{dt}{\ln (x)t-1}=\frac{dx}{x}\\[/tex]
still can't beak it as one type of variable on each side
so i substitute by another variable
[tex]z=\ln x\\[/tex]
[tex]dz=\frac{dx}{x}\\[/tex]
[tex]\frac{dt}{zt-1}=dz\\[/tex]
i don't know how to separate each variable type on one side
so i could integrate
??