Question

If 2x³ –3x²y² + 4x – y + 7 = 0 and y(1) = 1, then the value of y′′(1) is equal to-

• A. – $\frac{343}{474}$
• B. – $\frac{474}{343}$
• C. $\frac{474}{343}$
• D. $\frac{343}{474}$

Answer 1

Answer: (B)

– $\frac{474}{343}$

Answer 2

Sol. (1) y (1) = 1 given

⇒ x = 1, y = 1

(2) diff. 6x² –6xy² – 6x²yy′ + 4 – y′ = 0

put x = 1; y = 1

6 – 6 – 6y′(1) + 4 – y′ (1) = 0

⇒ y′ (1) = $\frac{4}{7}$

(3) Differentiating 12x – 6y² –12 xyy′′ – 6x² (y′)² – 6x²yy′′

= 0

Put x = 1 ; y = 1; y′(1) = 1

y′′(1) = – $\frac{474}{343}$