Question

If \[y = \frac{ax + b}{x^2 + c}\] then (2xy₁ + y)y₃ = 

 

Options

• 3(xy₂ + y₁)y₂

• 3(xy₁ + y₂)y₂

• 3(xy₂ + y₁)y₁

• none of these

Answer 1

(a) 3(xy₂ + y₁)y₂

Here,

\[y = \frac{ax + b}{x^2 + c}\]
\[ \Rightarrow \left( x^2 + c \right)y = ax + b\]
\[\text { Diffferentiating w . r . t . x, we get }\]
\[2xy + \left( x^2 + c \right)\frac{dy}{dx} = a\]
\[\text { Diffferentiating w . r . t . x, we get }\]
\[2y + 2x y_1 + 2x y_1 + \left( x^2 + c \right) y_2 = 0\]
\[ \Rightarrow 2y + 4x y_1 + \left( x^2 + c \right) y_2 = 0\]
\[\text { Diffferentiating again w . r . t . x, we get }\]
\[2 y_1 + 4 y_1 + 4x y_2 + \left( x^2 + c \right) y_3 + 2x y_2 = 0\]
\[ \Rightarrow 6 y_1 + 6x y_2 + \left( x^2 + c \right) y_3 = 0\]
\[ \Rightarrow 6 y_1 + 6x y_2 + \left( \frac{- 2y - 4x y_1}{y_2} \right) y_3 = 0 \left[ \because 2y + 4x y_1 + \left( x^2 + c \right) y_2 = 0 \right]\]
\[ \Rightarrow 6 y_1 y_2 + 6x \left( y_2 \right)^2 - 2y - 4x y_1 y_3 = 0\]
\[ \Rightarrow 3 y_1 y_2 + 3x \left( y_2 \right)^2 - y - 2x y_1 y_3 = 0\]
\[ \Rightarrow \left( y_1 + x y_2 \right)3 y_2 = \left( 2x y_1 + y \right) y_3 \]