A) 4
B) 3
C) 2
D) 1
Correct Answer: B
Solution :
\[y-\left( x+2{{y}^{2}} \right)\frac{dy}{dx}=0\]\[\Rightarrow \]\[y=\left( x+2{{y}^{2}} \right)\frac{dy}{dx}\] \[\Rightarrow \]\[y\frac{dy}{dx}=x+2{{y}^{2}}\]\[\Rightarrow \]\[\frac{dy}{dx}+\left( -\frac{1}{2} \right)x=2y\] \[I.F.={{e}^{\int_{{}}^{{}}{-\frac{1}{y}dy}}}={{e}^{-\ell ny}}={{y}^{-1}}={}^{1}/{}_{y}\] \[\therefore \]solution is \[x\left( \frac{1}{y} \right)=\int_{{}}^{{}}{\left( 2y \right)}\times \frac{1}{y}dy+c\] \[\Rightarrow \]\[\frac{x}{y}=2y+c\] \[x=1,y=-1\]\[\Rightarrow \]\[c=1\] \[\frac{x}{y}=2y+1\] put\[y=1\] \[x=2+1\] \[=3\]
You need to login to perform this action.
You will be redirected in
3 sec
