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Question

(A) Given the differential equation: $\frac{dy}{dx} = (x+y)^2$ $(i)$ Let $x+y = t$. Differentiating with respect to $x$,we get $1 + \frac{dy}{dx} = \f

Vedclass · Accepted Answer

$	an^{-1}(x+y) = x+C$

Vedclass · Answer

(A) Given the differential equation: $\frac{dy}{dx} = (x+y)^2$ $(i)$ Let $x+y = t$. Differentiating with respect to $x$,we get $1 + \frac{dy}{dx} = \frac{dt}{dx}$,which implies $\frac{dy}{dx} = \frac{dt}{dx} - 1$. Substituting this into equation $(i)$: $\frac{dt}{dx} - 1 = t^2$ $\frac{dt}{dx} = t^2 + 1$ Separating the variables: $\frac{dt}{t^2 + 1} = dx$ Integrating both sides: $\int \frac{dt}{t^2 + 1} = \int dx$ $	an^{-1}(t) = x + C$ Substituting $t = x+y$ back: $	an^{-1}(x+y) = x + C$

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

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