The solution of the equation frac{dy}{dx} = ( | vedclass

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(C) Given the differential equation $\frac{dy}{dx} = (x + y)^2$.Let $v = x + y$.Differentiating both sides with respect to $x$,we get $\frac{dv}{dx} =

Vedclass · Accepted Answer

$x + y - 	an(x + c) = 0$

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(C) Given the differential equation $\frac{dy}{dx} = (x + y)^2$.Let $v = x + y$.Differentiating both sides with respect to $x$,we get $\frac{dv}{dx} = 1 + \frac{dy}{dx}$,which implies $\frac{dy}{dx} = \frac{dv}{dx} - 1$.Substituting these into the original equation,we get $\frac{dv}{dx} - 1 = v^2$.Rearranging the terms,we have $\frac{dv}{dx} = v^2 + 1$.Separating the variables,we get $\frac{dv}{v^2 + 1} = dx$.Integrating both sides,we get $\int \frac{dv}{v^2 + 1} = \int dx$.This results in $	an^{-1}(v) = x + c$,where $c$ is the constant of integration.Substituting $v = x + y$ back,we get $	an^{-1}(x + y) = x + c$.Taking the tangent of both sides,we get $x + y = 	an(x + c)$,which can be rewritten as $x + y - 	an(x + c) = 0$.

The solution of the equation $\frac{dy}{dx} = (x + y)^2$ is

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