Consider the family of curves $A x^{2} + B y^{2} = 1$ . If this family of curves is represented by the differential equation $P y " + Q {(y')}^{2} = y y'$
, where P and Q are functions of x and y, here $(y' = \frac{d y}{d x} a n d y " = \frac{d^{2} y}{d x^{2}})$ then which of the following statement(s) is/are true ?

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$P = x y$

$P + Q = x (y + 1)$

$P - Q = y (x - 1)$

$Q = x y$

answer is A, C.

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Detailed Solution

$\begin{array}{l} A x^{2} + B y^{2} = 1 \Rightarrow A = \frac{1 - B y^{2}}{x^{2}} \\ Differentiate both sides \\ 0 = x^{2} (- 2 B y y') - (1 - B y^{2}) \cdot 2 x \\ 1 - B y^{2} = - B y x y' \Rightarrow B = \frac{1}{y^{2} - y x y'} \\ \frac{1}{B} = y^{2} - y x y' \Rightarrow 2 y y' - y x y " - y y' - x {(y')}^{2} = 0 \\ x {(y')}^{2} - y y' + y x y " = 0 \\ x y y " + x {(y')}^{2} = y y' \\ P = x y, Q = x \end{array}$