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## UnkleRhaukus 4 years ago Solve the following differential equation $(1+x^2)y''=1$It is y-absent

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1. anonymous

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

2. ash2326

( 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 $y= x \tan^{-1} x- 1/2( \ln (x^2+1)) + C1 x+C2$

3. UnkleRhaukus

THANKYOUS

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