A) \[y'''=\frac{3y'y'{{'}^{2}}}{1+y{{'}^{2}}}\]
B) \[y'''=\frac{3y'{{'}^{2}}}{1+y{{'}^{2}}}\]
C) \[y'''=\frac{3y'}{1+y{{'}^{2}}}\]
D) \[y'''=\frac{3y'}{1-y{{'}^{2}}}\]
Correct Answer: A
Solution :
[a] To eliminate the parameters g, f and c differentiate thrice w.r.t. x, \[x+yy'+g+fy'=0\] ? (1) \[1+y{{'}^{2}}+yy''+fy''=0\] ? (2) \[3y'y''+yy'''+fy'''=0\] ? (3) (1) \[y'''-(2)y''\] gives,You need to login to perform this action.
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