A) \[x+y+2=C{{e}^{y}}\]
B) \[x+y+4=C\,log\,y\]
C) \[\log (x+y+2)=Cy\]
D) \[\log (x+y+2)=C-y\]
Correct Answer: A
Solution :
[a] Putting \[x+y+1=u,\] we have \[du=dx+dy\]and the given equation reduces to \[u(du-dx)=dx\] \[\Rightarrow \frac{udu}{u+1}=dx\Rightarrow u-\log (u+1)=x\] \[\Rightarrow \log (x+y+2)=y+constant\Rightarrow x+y+2=C{{e}^{v}}\]You need to login to perform this action.
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