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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. rivermaker
    • 4 years ago
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    \[(1+x^2)y''=1 \rightarrow y'' = 1/(1+x^2)\] So just integrate twice

  2. ash2326
    • 4 years ago
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    ( 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
    • 4 years ago
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    THANKYOUS

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