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:
$\text{xy} = \text{log}\ \text{y} + \text{C} \ \ : \ \text{y}'=\frac{\text{y}^2}{1-\text{xy}}\ (\text{xy} \neq1)$

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Answer

Given: xy = log y + C ....(i) To prove: y given by eq. (i) is a solution of differential equation $\text{y}'=\frac{\text{y}^2}{1-\text{xy}} \ ...(\text{ii})$ Proof: Differentiating both sides of eq. (i) w.r.t x, we have $\text{xy}'+\text{y}(1)=\frac{1}{\text{y}}\text{y}'+0 \ $ $\Rightarrow \ \text{xy}'-\frac{\text{y}'}{\text{y}}=-\text{y} $ $ \Rightarrow \ \text{y}'\bigg(\text{x}-\frac{\text{1}}{\text{y}}\bigg)=-\text{y}$ $\Rightarrow \ \text{y}'\bigg(\frac{\text{xy}-1}{\text{y}}\bigg)=-\text{y}\ $ $\Rightarrow \ \text{y}'(\text{xy}-1)=-\text{y}^2$ $ \Rightarrow \ \text{y}'=\frac{-\text{y}^2}{\text{xy}-1}$$\Rightarrow \ \ \text{y}'=\frac{-\text{y}^2}{-(1-\text{xy})}=\frac{\text{y}^2}{1-\text{xy}}$
Hence, Function (implicit) given by eq. (i) is a solution of ${\text{y}}'=\frac{\text{y}^2}{1-\text{xy}}.$

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