The differential equation frac{dy}{dx} = frac | vedclass

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Question

(C) The given differential equation is $\frac{dy}{dx} = \frac{x+y}{1+x^2}$.We can rewrite this as $\frac{dy}{dx} = \frac{x}{1+x^2} + \frac{y}{1+x^2}$.

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(C) The given differential equation is $\frac{dy}{dx} = \frac{x+y}{1+x^2}$.We can rewrite this as $\frac{dy}{dx} = \frac{x}{1+x^2} + \frac{y}{1+x^2}$.Rearranging the terms,we get $\frac{dy}{dx} - \left(\frac{1}{1+x^2}\right)y = \frac{x}{1+x^2}$.This is in the standard form of a linear differential equation,$\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\frac{1}{1+x^2}$ and $Q(x) = \frac{x}{1+x^2}$.Therefore,the correct option is $C$.

The differential equation $\frac{dy}{dx} = \frac{x+y}{1+x^2}$ is a . . . . . . differential equation.

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