Verify that the function x + y = ^ -1 y (explicit or implicit) is… - Vidyadip

Question

Verify that the function $x + y = \tan^{-1}y ($explicit or implicit$)$ is a solution of differential equation $y^2y' + y^2 + 1 = 0.$

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Answer

Given: $x + y = \tan^{-1}y …(i)$
To prove:y given by eq. $(i)$ is a solution of differential equation $y^2y' + y^2 + 1 = 0 …(ii)$
Proof: Differentiating both sides of eq. $(i)$ w.r.t x we have
$1 + y' = \frac{1}{{1 + {y^2}}}y'$
$ \Rightarrow \left( {1 + y'} \right)\left( {1 + {y^2}} \right) = y'$
$ \Rightarrow 1 + {y^2} + y' + y'{y^2} = y'$
$ \Rightarrow {y^2}y' + {y^2} + 1 = 0$
Hence, function given by eq. $(i)$ is a solution of $y^2y' + y^2 + 1 = 0$

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