Question

(x+y)*dy/dx=1

Answer 1

The general solution to the differential equation is $x + y + 1 = C e^y$.

Answer 2

1. Rewrite the given differential equation $(x+y)\frac{dy}{dx}=1$ as $\frac{dx}{dy} = x+y$.
2. Rearrange it into the standard first-order linear differential equation form in $x$ as a function of $y$: $\frac{dx}{dy} - x = y$. Here, $P(y) = -1$ and $Q(y) = y$.
3. Calculate the Integrating Factor (IF): $IF = e^{\int P(y) dy} = e^{\int (-1) dy} = e^{-y}$
4. Multiply the linear equation by the IF: $e^{-y}\left(\frac{dx}{dy} - x\right) = y e^{-y}$ The left side is the derivative of the product of $x$ and the IF: $\frac{d}{dy}(x e^{-y}) = y e^{-y}$
5. Integrate both sides with respect to $y$: $\int \frac{d}{dy}(x e^{-y}) dy = \int y e^{-y} dy$ The left side is $x e^{-y}$. For the right side, use integration by parts ($\int u dv = uv - \int v du$) with $u=y$ and $dv=e^{-y}dy$. This yields $\int y e^{-y} dy = -y e^{-y} - e^{-y}$.
6. Equate the results and solve for $x$: $x e^{-y} = -y e^{-y} - e^{-y} + C$ where $C$ is the constant of integration. Multiply both sides by $e^y$: $x = -y - 1 + C e^y$
7. The general solution can be written as: $x + y + 1 = C e^y$