Substitution in Differential Equation

Knowing that $y=\tan { z }$ , differentiating it w.r.t. x ,

$\frac { dy }{ dx } =\sec ^{ 2 }{ z } \frac { dz }{ dx }$

Differentiating it again w.r.t. x ,

$\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =2\sec ^{ 2 }{ z } \tan { z } { \left( \frac { dz }{ dx } \right) }^{ 2 }+\sec ^{ 2 }{ z } \frac { { d }^{ 2 }z }{ d{ x }^{ 2 } }$

$\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =\sec ^{ 2 }{ z } \left[ 2\tan { z } { \left( \frac { dz }{ dx } \right) }^{ 2 }+\frac { { d }^{ 2 }z }{ d{ x }^{ 2 } } \right]$

Substituting into the original equation,

$\sec ^{ 2 }{ z } \left[ 2\tan { z } { \left( \frac { dz }{ dx } \right) }^{ 2 }+\frac { { d }^{ 2 }z }{ d{ x }^{ 2 } } \right] =1+\frac { 2\left( 1+\tan { z } \right) }{ 1+\tan ^{ 2 }{ z } } { \left( \sec ^{ 2 }{ z } \frac { dz }{ dx } \right) }^{ 2 }$

Simplifying the terms,

$2\tan { z } { \left( \frac { dz }{ dx } \right) }^{ 2 }+\frac { { d }^{ 2 }z }{ d{ x }^{ 2 } } =\cos ^{ 2 }{ z } +\frac { 2\left( 1+\tan { z } \right) }{ \sec ^{ 2 }{ z } } \sec ^{ 4 }{ z } { \left( \frac { dz }{ dx } \right) }^{ 2 }\frac { 1 }{ \sec ^{ 2 }{ z } }$

$2\tan { z } { \left( \frac { dz }{ dx } \right) }^{ 2 }+\frac { { d }^{ 2 }z }{ d{ x }^{ 2 } } =\cos ^{ 2 }{ z } +2\left( 1+\tan { z } \right) { \left( \frac { dz }{ dx } \right) }^{ 2 }$

$\frac { { d }^{ 2 }z }{ d{ x }^{ 2 } } =\cos ^{ 2 }{ z } +2\left( 1+\tan { z } -\tan { z } \right) { \left( \frac { dz }{ dx } \right) }^{ 2 }$

Hence, $\frac { { d }^{ 2 }z }{ d{ x }^{ 2 } } =\cos ^{ 2 }{ z } +2{ \left( \frac { dz }{ dx } \right) }^{ 2 }$

$\boxed{k = 2}$