Question

Solve the following differential equation
\[(1+x^2)y''=1\]It is y-absent

Answer 1

\[(1+x^2)y''=1 \rightarrow y'' = 1/(1+x^2)\]
So just integrate twice

Answer 2

( 1+x^2) y''=1
y''= 1/(+x^2)
\[y'=\tan^{-1} x + C1\]
C1 is constant of integration
integrating above with respect to x
\[y= x \tan^{-1} x- 1/2( \ln (x^2+1)) + C1 x+C2\]

Answer 3

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