The differential equation whose solution is $A x^2+B y^2=1$ where A and B are arbitrary constant is of: (A) first order and first degree Choose the correct answer from the options given below: |
(D) Only (C) and (D) Only (B) and (D) Only (A) Only |
(B) and (D) Only |
The correct answer is Option (3) - (B) and (D) Only order = no. of arbitrary constants $Ax^2+By^2=1$ differentiating wrt x $2Ax+2By\frac{dy}{dx}=0⇒-Ax=By\frac{dy}{dx}$ so $\frac{y}{x}\frac{dy}{dx}=-A$ again differentiating $\frac{y}{x}\frac{d^2y}{dx^2}+\frac{x\frac{dy}{dx}y}{x^2}\frac{dy}{dx}=0$ order = 2, degree = 1 |