question_answer

Solve the differential equation \[\left( \frac{{{e}^{-2\sqrt{x}}}}{\sqrt{x}}-\frac{y}{\sqrt{x}} \right)\,\,\frac{dx}{dy}=1,\] \[x\ne 0.\]

Answer:

We have, \[\left( \frac{{{e}^{-\,2\sqrt{x}}}}{\sqrt{x}}-\frac{y}{\sqrt{x}} \right)\frac{dx}{dy}=1\,\,;x\ne 0\] \[\Rightarrow \]   \[\frac{dy}{dx}=\frac{{{e}^{-\,2\sqrt{x}}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\Rightarrow \frac{dy}{dx}+\frac{y}{\sqrt{x}}=\frac{{{e}^{-\,2}}\sqrt{x}}{\sqrt{x}}\] which is a linear differential equation of the form \[\frac{dy}{dx}+Py=Q\]. \[P=\frac{1}{\sqrt{x}}\]and \[Q=\frac{{{e}^{-\,2\sqrt{x}}}}{\sqrt{x}}\]. \[\therefore \]      \[lF={{e}^{\int{pdx}}}={{e}^{\int{\frac{1}{\sqrt{x}}dx}}}={{e}^{2\sqrt{x}}}\] On multiplying both sides of Eq. (i) by \[lF={{e}^{2\sqrt{x}}}\], we get \[\frac{dy}{dx}{{e}^{2\sqrt{x}}}+\frac{y{{e}^{2\sqrt{x}}}}{\sqrt{x}}=\frac{1}{\sqrt{x}}\] On integrating both sides, we get \[\therefore \] \[\Rightarrow \]   \[y{{e}^{2\sqrt{x}}}=\int{\frac{1}{\sqrt{x}}\,dx+C}\] \[\Rightarrow \]   \[y{{e}^{2\sqrt{x}}}=2\sqrt{x}+C\] \[\Rightarrow \]   \[y=2\sqrt{x}+C{{e}^{-\,2}}\sqrt{x}\] which is the required solution of the given differential equation..