Solve the initial value problem. \[\frac{{{\bf{dy}}}}{{{\bf{dx}}}}{\bf{ = }}{{\bf{x}}^{\bf{2}}}\left( {{\bf{1 + y}}} \right){\bf{,}}\,\,\,\,\,\,\,\,\,\,\,\,{\bf{y}}\left( {\bf{0}} \right){\bf{ = 3}}\]

Given \(\frac{{{\bf{dy}}}}{{{\bf{dx}}\,}}{\bf{ = }}{{\bf{x}}^{\bf{2}}}\left( {{\bf{1 + y}}} \right)\, \cdot \cdot \cdot \cdot \cdot \cdot \left( {\bf{1}} \right)\)

Cross multiplication on both sides in the equation \(\left( 1 \right)\)

\[\frac{{{\bf{dy}}}}{{{\bf{1 + y}}}}{\bf{ = }}{{\bf{x}}^{\bf{2}}}\;{\bf{dx}}\,\,\]

Taking integration of both sides in the above equation

\(\int {\frac{{\bf{1}}}{{{\bf{1 + y}}}}} {\bf{dy = }}\int {{{\bf{x}}^{\bf{2}}}} \;{\bf{dx}}\,\,\)

Solve the above equation,

\[{\bf{ln}}\left( {{\bf{1 + y}}} \right){\bf{ = }}\frac{{{{\bf{x}}^{\bf{3}}}}}{{\bf{3}}}{\bf{ + c}}\, \cdot \cdot \cdot \cdot \cdot \cdot \left( {\bf{2}} \right)\]

Here c is the integration constant.

Substituting\(y\left( 0 \right) = 3\)in the equation\(\left( 2 \right)\)

\(\)\[\begin{array}{c}{\bf{ln}}\left( {{\bf{1 + 3}}} \right){\bf{ = }}\frac{{{{\left( {\bf{0}} \right)}^{\bf{3}}}}}{{\bf{3}}}{\bf{ + c}}\,\\{\bf{ln}}\left( {\bf{4}} \right){\bf{ = c}}\,\,\,\,\end{array}\]

Put the value of c in the equation\(\left( 2 \right)\)

\[{\bf{ln}}\left( {{\bf{1 + y}}} \right){\bf{ = }}\frac{{{{\bf{x}}^{\bf{3}}}}}{{\bf{3}}}{\bf{ + ln}}\left( {\bf{4}} \right)\]

Simplify the above equation,

\[\begin{array}{c}{\bf{ln}}\left( {{\bf{1 + y}}} \right){\bf{ = }}\frac{{{{\bf{x}}^{\bf{3}}}}}{{\bf{3}}}{\bf{ + ln}}\left( {\bf{4}} \right)\\{\bf{ln}}\left( {{\bf{1 + y}}} \right){\bf{ - ln}}\left( {\bf{4}} \right){\bf{ = }}\frac{{{{\bf{x}}^{\bf{3}}}}}{{\bf{3}}}\\{\bf{ln}}\left( {\frac{{{\bf{1 + y}}}}{{\bf{4}}}} \right){\bf{ = }}\frac{{{{\bf{x}}^{\bf{3}}}}}{{\bf{3}}}\\\frac{{{\bf{1 + y}}}}{{\bf{4}}}{\bf{ = }}{{\bf{e}}^{\frac{{{{\bf{x}}^{\bf{3}}}}}{{\bf{3}}}}}\\{\bf{y + 1 = 4}}{{\bf{e}}^{\frac{{{{\bf{x}}^{\bf{3}}}}}{{\bf{3}}}}}\\{\bf{y = 4}}{{\bf{e}}^{\frac{{{{\bf{x}}^{\bf{3}}}}}{{\bf{3}}}}}{\bf{ - 1}}\end{array}\]

Hence the solution is\[{\bf{y = 4}}{{\bf{e}}^{\frac{{{{\bf{x}}^{\bf{3}}}}}{{\bf{3}}}}}{\bf{ - 1}}\]