Question

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

A
first order and second degree
B
first order and first degree
C
second order and first degree
D
second order and second degree

Solution

## The correct option is C second order and first degreeDifferentiate the given equations for 2 times and eliminate A, B Ax2+By2=1 Differentiating 2Ax+2By−dydx=0 ⇒Ax+By.dydx=0                  - (1) Differentiating again A+B[y.d2ydx2+(dydx)2]=0            - (2) On solving (1) & (2); we will get values of A and B in terms of dydx,d2ydx2 & (dydx)2 and we insert those values in the given expression. Hence order = 2 Degree = 1Co-Curriculars

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