The general solution of the differential equa | vedclass

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Question

(A) Given: $(x + y) \frac{dy}{dx} = 1$. Rewrite as: $\frac{dx}{dy} = x + y$. This is a linear differential equation of the form $\frac{dx}{dy} - x = y

Vedclass · Accepted Answer

$x + y + 1 = ce^y$, where $c$ is a constant of integration

Vedclass · Answer

(A) Given: $(x + y) \frac{dy}{dx} = 1$. Rewrite as: $\frac{dx}{dy} = x + y$. This is a linear differential equation of the form $\frac{dx}{dy} - x = y$. Here, $P = -1$ and $Q = y$. The integrating factor $IF = e^{\int P dy} = e^{\int -1 dy} = e^{-y}$. The solution is given by $x \cdot IF = \int (Q \cdot IF) dy + c$. $x e^{-y} = \int y e^{-y} dy + c$. Using integration by parts: $\int y e^{-y} dy = y(-e^{-y}) - \int 1(-e^{-y}) dy = -y e^{-y} - e^{-y}$. So, $x e^{-y} = -y e^{-y} - e^{-y} + c$. Multiply by $e^y$: $x = -y - 1 + ce^y$. Rearranging gives: $x + y + 1 = ce^y$.

The general solution of the differential equation $(x + y) \frac{dy}{dx} = 1$ is

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