Sponsor Area

Continuity And Differentiability

Question
CBSEENMA12036209

The differential equation whose solution is Ax2 + By2 = 1, where A and B are arbitrary constants is of

  • second order and second degree

  • first order and second degree

  • first order and first degree

  • second order and first degree

Solution

D.

second order and first degree

Ax squared space plus By squared space equals space 1 space space space space..... space left parenthesis 1 right parenthesis
Ax space plus space By dy over dx space equals 0 space.... space left parenthesis 2 right parenthesis
straight A space plus By space fraction numerator straight d squared straight y over denominator dx squared end fraction space plus straight B open parentheses dy over dx close parentheses squared space equals 0 space.... space left parenthesis 3 space right parenthesis
straight x open curly brackets negative By fraction numerator straight d squared straight y over denominator dx squared end fraction minus straight B open parentheses dy over dx close parentheses squared close curly brackets plus By dy over dx space equals space 0
rightwards double arrow xy fraction numerator straight d squared straight y over denominator dx squared end fraction space plus straight x space open parentheses dy over dx close parentheses squared minus straight y dy over dx space equals space 0

Some More Questions From Continuity and Differentiability Chapter