A) \[{{y}^{2}}=4{{e}^{2x}}-2{{e}^{2x}}\,\,\ell ny\]
B) \[{{y}^{2}}={{e}^{2x}}+2{{e}^{2x}}\,\,\ell ny\]
C) \[{{y}^{2}}={{e}^{2x}}-\frac{1}{2}{{e}^{2x}}\,\,\ell ny\]
D) \[{{y}^{2}}={{e}^{2x}}+\frac{1}{2}{{e}^{2x}}\,\,\ell ny\]
Correct Answer: A
Solution :
\[\frac{dx}{dy}=\frac{{{e}^{2x}}+{{y}^{2}}}{{{y}^{3}}}\Rightarrow {{e}^{-2x}}\frac{dx}{dy}=\frac{1}{{{y}^{3}}}+\frac{1}{y}{{e}^{-2x}}\] Put \[-{{e}^{-2x}}=t\Rightarrow \frac{dt}{dy}+\frac{2t}{y}=\frac{2}{{{y}^{3}}}\] (Linear differential equation in t) \[\therefore \,\,\,\,t.{{y}^{2}}=\int{\frac{2}{{{y}^{3}}}}.{{y}^{2}}dy+C\] \[\Rightarrow \,-{{e}^{-2x}}.{{y}^{2}}=2\ell n\,y+C.\]Now \[y(0)=1\] \[\Rightarrow \,C=-1\] \[\therefore \,\,-{{e}^{-2x}}.{{y}^{2}}=2\ell n\,\,y-1\] \[\Rightarrow \,\,{{y}^{2}}={{e}^{2x}}-2{{e}^{2x}}\ell n\,\,y\]
You need to login to perform this action.
You will be redirected in
3 sec
