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

Question

Verify that the function x + y = tan^-1y (explicit or implicit) is a solution of differential equation y²y' + y² + 1 = 0.

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Answer

Given: x + y = tan^-1y …(i)
To prove:y given by eq. (i) is a solution of differential equation y²y' + y² + 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 y²y' + y² + 1 = 0

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