If y-tan-1-1-a2x2-1ax- then 1-a2x2y-2a2xy1 -

The correct option is D 0 y=tan^ 1[√(1+a^2x^2) 1ax] Put ax=tanθ =tan^ 1[√(1+tan^2θ) 1tanθ] =tan^ 1[θ 1tanθ] =tan^ 1[1cosθ 1sinθco ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Integration of Trigonometric Functions

If y=tan-1√...

Question

If $y = {tan}^{- 1} (\frac{\sqrt{1 + a^{2} x^{2}} - 1}{a x}),$ then $(1 + a^{2} x^{2}) y^{''} + 2 a^{2} x y^{1} =$

- 2 a^{2}

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

a^{2}

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

2 a^{2}

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

0

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

Open in App

Solution

Suggest Corrections

Similar questions

x = \frac{1}{a_{1} +} \frac{a}{a_{2} +} \frac{1}{a_{1} +} \frac{1}{a_{2} +} . . . .,

y = \frac{1}{2 a_{1} +} \frac{1}{2 a_{2} +} \frac{1}{2 a_{1} +} \frac{1}{2 a_{2} +} . . . . . .,

z = \frac{1}{3 a_{1} +} \frac{1}{3 a_{2} +} \frac{1}{3 a_{1} +} \frac{1}{3 a - 2 +} . . . .,

x (y^{2} - z^{2}) + 2 y (z^{2} - x^{2}) + 3 z (x^{2} - y^{2}) = 0

If X = 1 - a^{2} and Y = 1 - b^{2}, then what is X^{2} + Y^{2} ?

{(1 + x + x^{2} + \cdot \cdot + x^{n})}^{n}

= a_{0} + a_{1} x + a_{2} x^{2} + \cdot \cdot \cdot + a_{m p} x^{m p}

a_{1} + 2 a_{2} + 3 a_{3} + \cdot \cdot \cdot + n p a_{n p}

(1 + x + x^{2} + . . . . . + x^{p})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . a_{n p} x^{n p}

(1) a_{0} + a_{1} + a_{2} + . . . . . + a_{n p} = (p + 1)^{n}

(2) a_{1} + 2 a_{2} + 3 a_{3} + . . . . + n p . a_{n p} = \frac{1}{2} n p (p + 1)^{n}

{(a + \frac{1}{2 a +} \frac{1}{2 a +} \dots)}^{2} + {(a - \frac{1}{2 a -} \frac{1}{2 a -} \dots)}^{2} = 2 a^{2}

(a + \frac{1}{2 a +} \frac{1}{2 a +} \dots) (a - \frac{1}{2 a -} \frac{1}{2 a -} \dots) = a^{2} - \frac{1}{2 a^{2} -} \frac{1}{2 a^{2} -} \dots

Join BYJU'S Learning Program