Question

The general solution of the differential equation $(x + y + 3)\frac{dy}{dx} = 1$ is:

• A. $x + y + 3 = Ce^y$
• B. $x + y + 4 = Ce^y$
• C. $x + y + 3 = Ce^{-y}$
• D. $x + y + 4 = Ce^{-y}$
• E. $x + y + 4e^y = C$

Hint:

When $\frac{dy}{dx}$ results in an "inseparable" mess in the denominator, always try flipping it to $\frac{dx}{dy}$. Many non-linear $y$-focused equations are actually simple linear $x$-focused equations.

Answer 1

The Correct Option is B

Concept: This differential equation can be treated as a linear differential equation if we invert the derivative to find $\frac{dx}{dy}$. This transforms it into the form $\frac{dx}{dy} + P(y)x = Q(y)$.

Step 1: Rearrange to find $\frac{dx}{dy}$.
\[ \frac{dx}{dy} = x + y + 3 \quad \Rightarrow \quad \frac{dx}{dy} - x = y + 3 \] This is a linear differential equation in $x$ where $P(y) = -1$ and $Q(y) = y + 3$.

Step 2: Find the Integrating Factor (I.F.).
\[ \text{I.F.} = e^{\int P(y) \, dy} = e^{\int -1 \, dy} = e^{-y} \]

Step 3: Solve the equation.
\[ x \cdot (\text{I.F.}) = \int Q(y) \cdot (\text{I.F.}) \, dy \] \[ x e^{-y} = \int (y + 3) e^{-y} \, dy \] Integrate by parts: $\int u v \, dy = u \int v \, dy - \int (u' \int v \, dy) \, dy$ \[ \int (y + 3) e^{-y} \, dy = (y + 3)(-e^{-y}) - \int (1)(-e^{-y}) \, dy \] \[ = -(y + 3)e^{-y} - e^{-y} + C = e^{-y}(-y - 3 - 1) + C = e^{-y}(-y - 4) + C \] \[ x e^{-y} = - (y + 4)e^{-y} + C \] Multiply through by $e^y$: \[ x = -(y + 4) + Ce^y \quad \Rightarrow \quad x + y + 4 = Ce^y \]