Question

The differential equation satisfied by ax² + by² = 1 is

पर्याय

• xyy₂ + y₁² + yy₁ = 0

• xyy₂ + xy₁² − yy₁ = 0

• xyy₂ − xy₁² + yy₁ = 0

• none of these

Answer 1

xyy₂ + xy₁² − yy₁ = 0

 

We have,
ax² + by² = 1                                    .....(1)
Differentiating both sides of (1) with respect to x, we get
\[2ax + 2by\frac{dy}{dx} = 0 . . . . . \left( 2 \right)\]
Differentiating both sides of (2) with respect to x, we get
\[2a + 2b \left( \frac{dy}{dx} \right)^2 + 2by\frac{d^2 y}{d x^2} = 0\]
\[ \Rightarrow 2b\left[ y\frac{d^2 y}{d x^2} + \left( \frac{dy}{dx} \right)^2 \right] = - 2a\]
\[ \Rightarrow \left[ y\frac{d^2 y}{d x^2} + \left( \frac{dy}{dx} \right)^2 \right] = - \frac{2a}{2b}\]
\[ \Rightarrow \left[ y\frac{d^2 y}{d x^2} + \left( \frac{dy}{dx} \right)^2 \right] = - \left( - \frac{y}{x}\frac{dy}{dx} \right) .............\left[\text{Using (2)}\right]\]
\[ \Rightarrow x\left[ y\frac{d^2 y}{d x^2} + \left( \frac{dy}{dx} \right)^2 \right] = y\frac{dy}{dx}\]
\[ \Rightarrow xy\frac{d^2 y}{d x^2} + x \left( \frac{dy}{dx} \right)^2 = y\frac{dy}{dx}\]
\[ \Rightarrow xy\frac{d^2 y}{d x^2} + x \left( \frac{dy}{dx} \right)^2 - y\frac{dy}{dx} = 0\]
\[ \Rightarrow xy y_2 + x \left( y_1 \right)^2 - y y_1 = 0\]