A) 3 only
B) -1 only
C) Both -1 and 3
D) Neither -1 nor 3
Correct Answer: A
Solution :
| [a] Given, \[xdy=ydx+{{y}^{2}}dy\] |
| \[\Rightarrow 1=\frac{4}{x}\frac{dx}{dy}+\frac{{{y}^{2}}}{x}\Rightarrow \frac{dx}{dy}+y=\frac{x}{y}\] |
| \[\Rightarrow \frac{dx}{dy}=\frac{x}{y}=-y\] ? (1) |
| \[P=-\frac{1}{y},Q=-y\] |
| IF \[=\,\,{{e}^{\int\limits_{{}}{Pdy}}}=\,\,e\,{{\,}^{\int\limits_{e}{-\frac{1}{y}dy}}}\,\,=\,\,{{e}^{-\log y}}=\frac{1}{y}\] |
| Multiplying Eqn. (1) by IF |
| \[\Rightarrow \frac{1}{y}\frac{dx}{dy}-\frac{x}{{{y}^{2}}}=-1;\] \[\frac{x}{y}=\int{\frac{1}{y}(-y)dy+C}\] |
| \[\Rightarrow \frac{x}{y}=\int{-1dy+C\Rightarrow \frac{x}{y}=-y+C}\] |
| \[y(1)=1\] |
| \[\frac{1}{1}=-1+C\Rightarrow C=2\] |
| \[\Rightarrow \frac{x}{y}=-y+2\Rightarrow x=-{{y}^{2}}+2y\] |
| \[\Rightarrow y(-3)\Rightarrow -3=-{{y}^{2}}+2y\Rightarrow {{y}^{2}}-2y-3=0\] |
| \[\Rightarrow y=\frac{+2\pm \sqrt{4+12}}{2}=\frac{2\pm 4}{2}\Rightarrow y=3,-1\] |
| Since y>0 so y=3. |
You need to login to perform this action.
You will be redirected in
3 sec
