Show that an implicit solution of 2x sin2(y) dx − (x2 + 12) cos(y) dy = 0 is given by ln(x2 + 12) + csc(y) = C. Differentiating ln(x2 + 12) + csc(y) = C we get 2x x2 + 12 + dy dx = 0 or 2x sin2(y) dx + dy = 0. Find the constant solutions, if any, that were lost in the solution of the differential equation. (Let k represent an arbitrary integer.) y =

[tex]2x\sin (2y)dx=(x^2+12)\cos y dy\\ \\\dfrac{2x\sin (2y)dx}{x^2+12}=\cos ydy\ [\text{Divided by non-zero expression }x^2+12]\\ \\\dfrac{2x}{x^2+12}dx=\dfrac{\cos y}{\sin (2y)}dy\ [\text{Divided by }\sin (2y)]\\ \\\dfrac{2x}{x^2+12}dx=\dfrac{\cos y}{2\sin y\cos y}dy\ [\text{Use formula }\sin (2y)=2\sin y\cos y]\\ \\\dfrac{2x}{x^2+12}dx=\dfrac{1}{2\sin y}dy\ [\text{Simplify when }\cos y\neq 0][/tex]

Now,

[tex]\int \dfrac{2x}{x^2+12}dx=\int \dfrac{1}{2\sin y}dy\\ \\\int \dfrac{d(x^2)}{x^2+12}=\dfrac{1}{2}\int \dfrac{\sin y}{\sin^2 y}dy\\ \\\int \dfrac{d(x^2+12)}{x^2+12}=-\dfrac{1}{2}\int \dfrac{d(\cos y)}{1-\cos^2 y}\\ \\\ln (x^2+12)+C=-\dfrac{1}{2}\int \left(\dfrac{1}{2(1-\cos y)}+\dfrac{1}{2(1+\cos y)}\right)d(\cos y)\\ \\\ln (x^2+12)+C=-\dfrac{1}{4}\int \dfrac{d(\cos y)}{1-\cos y}-\dfrac{1}{4}\int \dfrac{d(\cos y)}{1+\cos y}\\ \\\ln (x^2+12)+C=\dfrac{1}{4}\ln (1-\cos y)-\dfrac{1}{4}\ln (1+\cos y)[/tex]

[tex]\ln (x^2+12)+C=\dfrac{1}{4}\ln \dfrac{1-\cos y}{1+\cos y}[/tex]

Find the constant solutions, if any, that were lost in the solution of the differential equation:

When

[tex]\cos y=0,[/tex]

then

[tex]y=\dfrac{\pi }{2}+\pi k,\ k\in Z[/tex]