In each of the verify that the given function (explicit or implic… - Vidyadip

Question

In each of the verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:
$\text{x + y} = \text{tan}^{–1}\ \text{y} \ \ :\ \text{y}^2 \text{y}' + \text{y}^2 + 1 = 0$

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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
$1+\text{y}'=\frac{1}{1+\text{y}^2}\text{y}' \ \Rightarrow \ \ (1+\text{y}')(1+\text{y}^2 ) = \text{y}'$
$\Rightarrow\ \ \ \ 1+ \text{y}^2 + \text{y}'+ \text{y}' \text{y}^2 = \text{y}' $ $\Rightarrow\ \text{y}^2 \text{y}'+ \text{y}^2 +1 -0 = \text{eq}. (\text{ii})$
Hence, Function given by eq. (i) is a solution of $y^2 y' + y^2 + 1 = 0$

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