If $y^{1 / m} + y^{- 1 / m} = 2 x$ then $(x^{2} - 1) y_{2} + x y_{1} =$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

-my

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- m^{2} y

No worries! We‘ve got your back. Try BYJU‘S free classes today!

m^{2} y

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $m^{2} y$
Applying derivative,
$\frac{y^{\frac{1}{m} - 1}}{m} y_{1} - \frac{y^{- \frac{1}{m} - 1}}{m} y_{1} = 2 ⎛ ⎝ y^{\frac{1}{m}} - y^{- \frac{1}{m}} ⎞ ⎠ = \frac{2 m y}{y_{1}}$
Now, ${⎛ ⎝ y^{\frac{1}{m}} - y^{- \frac{1}{m}} ⎞ ⎠}^{2} = {⎛ ⎝ y^{\frac{1}{m}} + y^{- \frac{1}{m}} ⎞ ⎠}^{2} - 4 ⟹ {(\frac{2 m y}{y_{1}})}^{2} = {(2 x)}^{2} - 4 ⟺ m^{2} y^{2} = (x^{2} - 1) y_{1}^{2}$
Appluying derivative,
$m^{2} (2 y y_{1}) = 2 x y_{1}^{2} + (x^{2} - 1) 2 y_{1} y_{2}$
Cancelling $2 y_{1} (A s s u m i n g \neq 0) ⟹ m^{2} y = x y_{1} + (x^{2} - 1) y_{2}$
Hence the answer is $m^{2} y$ .

Suggest Corrections

Similar questions

y^{\frac{1}{m}} + y^{\frac{- 1}{m}} = 2 x

(x^{2} - 1) y_{2} + x y_{1} - m^{2} y = 0

y^{1 / m} + y^{- 1 / m} = 2 x

(x^{2} - 1) y_{2} + x y_{1}

y^{1 / m} = [x + \sqrt{1 + x^{2}}]

(1 + x^{2}) y_{2} + x y_{1}

(x + \sqrt{x^{2} - 1}) m

(1 - x^{2}) y_{2} - x y_{1} + m^{2} y = 0

y = (x + \sqrt{1 + x^{2}})^{m}

(1 + x^{2}) y_{2} + x y_{1} - m^{2} y =

Join BYJU'S Learning Program