Question

The differential equation whose solution is $A{x}^2+B{y}^2=1$ 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

Answer 1

The correct option is C second order and first degree

Differentiate the given equations for 2 times and eliminate A, B
$A{x}^2+B{y}^2=1$
Differentiating
$2Ax+2By-\frac{dy}{dx}=0$
$\Rightarrow Ax+By.\frac{dy}{dx}=0$ - (1)
Differentiating again
$A+B[y.\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^2]=0$ - (2)
On solving (1) & (2); we will get values of A and B in terms of $\frac{dy}{dx},\frac{d^{2}y}{d{x}^2}$ & ${\left(\frac{dy}{dx}\right)}^2$ and we insert those values in the given expression.
Hence order = 2
Degree = 1