91影视

Given \(\sqrt {\bf{y}} {\bf{dx + }}\left( {{\bf{1 + x}}} \right){\bf{dy = 0}}\)

Simplify the above equation,

\(\sqrt y \;{\bf{dx = - }}\left( {{\bf{1 + x}}} \right){\bf{dy}}\, \cdot \cdot \cdot \cdot \cdot \cdot \left( {\bf{1}} \right)\)

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

\(\frac{{\bf{1}}}{{\left( {{\bf{1 + x}}} \right)}}{\bf{dx = }}\frac{{{\bf{ - 1}}}}{{\sqrt {\bf{y}} }}{\bf{dy}}\)

Taking integration of both sides in the above equation

\(\int {\frac{{\bf{1}}}{{\left( {{\bf{1 + x}}} \right)}}} {\bf{dx = - }}\int {\frac{{\bf{1}}}{{\sqrt {\bf{y}} }}} {\bf{dy}}\)

Solve the above equation,

\({\bf{ln}}\left( {{\bf{1 + x}}} \right){\bf{ = - ln}}\left( {\sqrt {\bf{y}} } \right){\bf{ + c}}\, \cdot \cdot \cdot \cdot \cdot \cdot \left( {\bf{2}} \right)\)

Here c is the integration constant.

Substituting\(y\left( 0 \right) = 1\)in the equation\(\left( 2 \right)\)and simplifying,

\(\begin{array}{c}{\bf{ln}}\left( {{\bf{1 + 0}}} \right){\bf{ = - ln}}\left( {\sqrt {\bf{1}} } \right){\bf{ + c}}\\{\bf{ln}}\left( {\bf{1}} \right){\bf{ = - ln}}\left( {\bf{1}} \right){\bf{ + c}}\\{\bf{c = 0}}\end{array}\)

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

\[\begin{array}{l}{\bf{ln}}\left( {{\bf{1 + x}}} \right){\bf{ = - ln}}\left( {\sqrt {\bf{y}} } \right){\bf{ + 0}}\\{\bf{ln}}\left( {{\bf{1 + x}}} \right){\bf{ = - ln}}\left( {\sqrt {\bf{y}} } \right)\end{array}\]

Simplify the above equation,

\(\begin{array}{c}{\bf{ln}}\left( {{\bf{1 + x}}} \right){\bf{ = - ln}}\left( {\sqrt {\bf{y}} } \right)\\{\bf{ln}}\left( {{\bf{1 + x}}} \right){\bf{ + ln}}\left( {\sqrt {\bf{y}} } \right){\bf{ = 0}}\\\sqrt {\bf{y}} \left( {{\bf{1 + x}}} \right){\bf{ = 0}}\end{array}\)

Hence the solution is\[\sqrt {\bf{y}} \left( {{\bf{1 + x}}} \right){\bf{ = 0}}\]