Step-by-step explanation:
We are given:
[tex]xy'=coty[/tex]
This can be rewritten as
[tex]x\frac{dy}{dx} =coty[/tex]
Next, we can bring the x's and y's to their respective sides by dividing by coty and x and then multiplying the dx to the other side. We can then change [tex]\frac{1}{coty}[/tex] into [tex]tany[/tex]. This gives us the differential
[tex]tany\,dy=\frac{1}{x} \,dx[/tex]
Now we can integrate each side
[tex]\int tan(y)\,dy=\int \frac{1}{x} \,dx[/tex]
To integrate tan(y), we need to manipulate it
[tex]\int \frac{sin(y)}{cos(y)} \,dy=\int \frac{1}{x} \,dx[/tex]
Now we can use u-substitution where [tex]u= cos(y)\\du=-sin(y) dy[/tex]
This gives us
[tex]-\int \frac{1}{u} \,du=\int \frac{1}{x} \,dx[/tex]
Now, lets integrate both sides
[tex]-ln|u|=ln|x|+c[/tex]
Next, we can substitute our u value back in
[tex]-ln|cos(y)|=ln|x|+c[/tex]
Now we can add [tex]-ln|cos(y)|[/tex] to the other side and subtract c from each side. This gives us
[tex]C_2=ln|x|+ln|cos(y)|[/tex]
Next, we can apply a property of logarithms to combine this sum of two logs into one log.
[tex]C_2=ln|xcos(y)|[/tex]
Lastly, we can add a base e to each side to remove the ln
[tex]C_3=|xcos(y)|[/tex]
And here is our answer.
