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【General guidance】The answer provided below has been developed in a clear step by step manner.Step1/2given diffrential equation is \( \mathrm{{\cos{{y}}}{\left({2}{\ln{{x}}}-{x}^{{3}}\right)}\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}=\frac{{2}}{{x}}-{3}{x}^{{2}}} \)now, we seprate x and y variable\( \begin{align*} \mathrm{{\cos{{y}}}.{\left.{d}{y}\right.}} &= \mathrm{\frac{{{\left(\frac{{2}}{{x}}-{3}{x}^{{2}}\right)}.{\left.{d}{x}\right.}}}{{{2}{\ln{{x}}}-{x}^{{3}}}}} \end{align*} \)Explanation:Please refer to solution in this step.Step2/2integrating both side, we get\( \begin{align*} \mathrm{\int{\cos{{y}}}{\left.{d}{y}\right.}} &= \mathrm{\int\frac{{{\left(\frac{{2}}{{x}}-{3}{x}^{{2}}\right)}{\left.{d}{x}\right.}}}{{{2}{\ln{{x}}}-{x}^{{3}}}}} \end{align*} \)put \( \mathrm{{2}{\ln{{x}}}-{x}^{{3}}={t}} \)then \( \mathrm{{\left(\frac{{2}}{{x}}-{3}{x}^{{2}}\right)}{\left.{d}{x}\right.}={\left.{d}{t}\right.}} \)so, above integral becomes\( \begin{align*} \mathrm{\int{\cos{{y}}}{\left.{d}{y}\right.}} &= \mathrm{\int\frac{{{\left.{d}{t}\right.}}}{{t}}} \\[3pt]\mathrm{{\sin{{y}}}} &= \mathrm{{\ln{{t}}}+{c}} \end{align*} \)putting the value of t in above, we get\( \begin{align*} \mathrm{{\sin{{y}}}} &= \mathrm{{\ln{{\left|{{2}{\ln{{x}}}-{x}^{{3}}}\right|}}}+{C};{C}={c}{o}{n}{s}{\tan{{t}}}} \\[3pt]\mathrm{{y}} &= \mathrm{{\arcsin{{\left({\ln{{\left|{{2}{\ln{{x}}}-{x}^{{3}}}\right|}}}+{C}\right)}}}} \end{align*} \)Explanation:\( \mathrm{\int{\cos{{x}}}{\left.{d}{x}\right.}={\sin{{x}}}+{c}{\quad\text{and}\quad}\int\frac{{{\left.{d}{x}\right.}}}{{x}}={\log{{x}}}+{C}} \), where C is integral constant. ...