Question

यदि y=(tan^{-1} x)^{2} है तो दर्शाइए कि (x^{2}+1)^{2} y_{2} +2x(x^{2}+1)y_{1}=2 है।

Answer 1

Given -----> y = ( tan⁻¹ x )²

To prove ----->

( x² + 1 )² y₂ + 2x ( x² + 1 ) y₁ = 2

proof-----> ATQ,

y = ( tan⁻¹ x )²

Differentiating with respect to x , we get,

=> dy / dx = d / dx ( tan⁻¹ x )²

We know that, dy / dx = y₁ , putting it , we get,

=> y₁ = 2 ( tan⁻¹ x )²⁻ ¹ d/dx ( tan⁻¹ x )

= 2 ( tan⁻¹ x ) { 1 / ( 1 + x² ) }

=> ( 1 + x² ) y₁ = 2 ( tan⁻¹ x )

Squaring both sides we get,

=> ( 1 + x² )² y₁² = 4 ( tan⁻¹ x )²

Now putting tan⁻¹ x = y , we get,

=> ( 1 + x² )² y₁² = 4 y

Differentiating with respect to x , we get,

=> d/dx { ( 1 + x² )² y₁² } = 4 dy/dx

=> ( 1 + x² )² d/dx ( y₁² ) + y₁² d/dx ( 1 + x² )² = 4y₁

=>(1 + x²)² 2 y₁²⁻¹ d/dx( y₁ )+ y₁² 2 (1 + x²)¹ d/dx (1+ x²)

= 4 y₁

=> 2 (1 + x² )² y₁ y₂ + 2 y₁² ( 1 + x² ) ( 0 + 2x ) =4 y₁

=> 2 y₁ { ( 1 + x² )² y₂ + 2x ( 1 + x² ) y₁ } = 4 y₁

2 y₁ is cancel out from both sides , and we get,

=> ( 1 + x² )² y₂ + 2 x ( 1 + x² ) y₁ = 2

Answer 2

$(x^{2}+1)^{2} y_{2} +2x(x^{2}+1)y_{1}=2$   यदि y= (Tan⁻¹x)²

Step-by-step explanation:

y  = (Tan⁻¹x)²

dy/dx = 2 (Tan⁻¹x) / (1 + x²)

dy/dx  = y₁

=>  (1 + x²)y₁ = 2 (Tan⁻¹x)

dy/dx = 2 (Tan⁻¹x) / (1 + x²)

d²y/dx²   = 2 (Tan⁻¹x)  (-2x/(1+ x²)²)   + 2/(1 + x²)(1 + x²)

d²y/dx²   =  y₂

=> y₂  = 2( 1  - 2x (Tan⁻¹x) )/(1+ x²)²

=> (1+ x²)² y₂ = 2 - 4x (Tan⁻¹x)

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

LHS

= 2 - 4x (Tan⁻¹x)  + 2x(2 (Tan⁻¹x))

= 2 - 4x (Tan⁻¹x)  + 4x (Tan⁻¹x)  

= 2

= RHS

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