Question

The differential equation whose general solution is $Ax^2 + By^2 = 1$, where $A$ and $B$ are arbitrary constants is of ?

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

Answer 1

The Correct Option is B

$Ax^2 + By^2 = 1$ can be written as
$By^2 = 1 -Ax^2$
Differentiating w.r.t. $'x'$, we get
$2By \frac{dy}{dx}=-2Ax$
$ \Rightarrow \frac{dy}{dx}=- \frac{A}{B} \frac{x}{y} \Rightarrow -\frac{A}{B} =\frac{y}{x} \frac{dy}{dx}$
Again differentiating w.r.t. $'x'$, we get
$\frac{d^{2}y}{dx^{2}} =\left(-\frac{A}{B} \right)\left(\frac{y-x \frac{dy}{dx}}{y^{2}}\right)$
$= \left(-\frac{A}{B}\right)\left(\frac{y-x\left(-\frac{A}{B}. \frac{x}{y}\right)}{y^{2}}\right)$
$\frac{d^{2}y}{dx^{2}}=\left(\frac{y}{x} . \frac{dy}{dx}\right) \frac{y-x\left(\frac{y}{x} . \frac{dy}{dx} . \frac{x}{y}\right)}{y^{2}}$
$=\frac{y}{x} . \frac{dy}{dx}\left(\frac{y-x \frac{dy}{dx}}{y^{2}}\right)$
$\Rightarrow y^{2} \frac{d^{2} y}{dx^{2}} =\frac{y^{2}}{x} \frac{dy}{dx} -y\left(\frac{dy}{dx}\right)^{2} $
$\Rightarrow y \frac{d^{2}y}{dx^{2}}+ \left(\frac{dy}{dx}\right)^{2} -\frac{y}{x} \frac{dy}{dx}= 0$
$ \Rightarrow xy \frac{d^{2}y}{dx^{2}} +x \left(\frac{dy}{dx}\right)^{2} -y \frac{dy}{dx}= 0$
has 2nd order and first degree.