Question:

Step 1: Expand the given differential equation. We start with $(1 + \tan y)(dx - dy) + 2x\, dy = 0$. Expanding: $$(1 + \tan y)dx - (1 + \tan y)dy + 2x\,dy = 0$$ Collecting terms with $dx$ and $dy$: $$(1 + \tan y)dx + [-(1+\tan y) + 2x]dy = 0$$ Rearranging: $$(1 + \tan y)dx = (1 + \tan y - 2x)dy$$ Step 2: Convert to a linear differential equation in $x$. Dividing both sides by $(1 + \tan y)dy$: $$\frac{dx}{dy} = \frac{1 + \tan y - 2x}{1 + \tan y}$$ Separating the right side: $$\frac{dx}{dy} = 1 - \frac{2x}{1+\tan y}$$ Rearranging into standard linear form: $$\frac{dx}{dy} + \frac{2x}{1+\tan y} = 1$$ Step 3: Find the integrating factor. This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$ where $P(y) = \frac{2}{1+\tan y}$ and $Q(y) = 1$. The integrating factor is: $$\mu(y) = e^{\int P(y)\,dy} = e^{\int \frac{2}{1+\tan y}dy}$$ Step 4: Simplify the exponent integral. Rewrite the integrand: $$\frac{2}{1+\tan y} = \frac{2\cos y}{\cos y + \sin y}$$ To integrate $\int \frac{2\cos y}{\sin y + \cos y}dy$, we use partial decomposition. Write: $$2\cos y = A(\sin y + \cos y) + B(\cos y - \sin y)$$ Expanding the right side: $$2\cos y = (A+B)\cos y + (A-B)\sin y$$ Comparing coefficients: - Coefficient of $\cos y$: $A + B = 2$ - Coefficient of $\sin y$: $A - B = 0$ Solving: $A = 1$ and $B = 1$. Therefore: $$\int \frac{2\cos y}{\sin y + \cos y}dy = \int \frac{(\sin y+\cos y)+(\cos y - \sin y)}{\sin y+\cos y}dy$$ $$= \int 1\,dy + \int \frac{\cos y - \sin y}{\sin y+\cos y}dy = y + \ln|\sin y + \cos y|$$ Step 5: Calculate the integrating factor. $$\mu(y) = e^{y + \ln|\sin y+\cos y|} = e^y \cdot e^{\ln(\sin y + \cos y)} = e^y(\sin y + \cos y)$$ Step 6: Multiply the differential equation by the integrating factor. Multiplying $\frac{dx}{dy} + \frac{2x}{1+\tan y} = 1$ by $\mu(y) = e^y(\sin y + \cos y)$: $$\frac{d}{dy}[x \cdot e^y(\sin y+\cos y)] = e^y(\sin y+\cos y)$$ Step 7: Integrate both sides with respect to $y$. $$x \cdot e^y(\sin y+\cos y) = \int e^y(\sin y+\cos y)\,dy$$ Using the standard result $\int e^y(\sin y+\cos y)\,dy = e^y \sin y + C$: $$x \cdot e^y(\sin y+\cos y) = e^y \sin y + C$$ Step 8: Solve for the general solution. Dividing both sides by $e^y$: $$x(\sin y+\cos y) = \sin y + Ce^{-y}$$ **Final Answer:** The general solution is $x(\sin y + \cos y) = \sin y + Ce^{-y}$, which corresponds to **Option 2**.