Answer to Question #193561 in Differential Equations for Nishona

Question

cotxdy=ydx=(cotx)(3e^sinx)dx

Answer 1

If cotxdy=ydx=(cotx)(3e^sinx)dx then $\left\{ \begin{array}{l} ydx = \cot xdy\\ ydx = \cot x \cdot 3{e^{\sin x}}dx \end{array} \right.$

From 1-st equation:

$\frac{{dy}}{y} = \frac{{\sin x}}{{\cos x}}dx = - \frac{{d\cos x}}{{\cos x}} \Rightarrow \ln y = - \ln \cos x + \ln {C_1} = \ln \frac{{{C_1}}}{{\cos x}} \Rightarrow y = \frac{{{C_1}}}{{\cos x}}$

From 2-nd equation:

$ydx = \cot x \cdot 3{e^{\sin x}}dx \Rightarrow \frac{{{C_1}}}{{\cos x}}dx = \cot x \cdot 3{e^{\sin x}}dx \Rightarrow \frac{{{C_1}\sin x}}{{\cos x}}dx = \cos x \cdot 3{e^{\sin x}}dx \Rightarrow - {C_1}\frac{{d\cos x}}{{\cos x}} = 3{e^{\sin x}}d\sin x \Rightarrow - {C_1}\ln \cos x = 3{e^{\sin x}} + {C_2} \Rightarrow {C_1}\ln \frac{1}{{\cos x}} = 3{e^{\sin x}} + {C_2}$

The resulting equality will not hold for any ${C_1}, {C_2}$ .

So, system has no solutions.

Answer: There are no solutions.