Question

Transform $dx + xdy = e^{-y} \sec^2 y \, dy$ into linear form

• A. $\frac{dx}{dy} - x = e^{-y} \sec^2 y$
• B. $\frac{dx}{dy} = e^{-y} \sec^2 y$
• C. $\frac{dx}{dy} + x = e^{-y} \sec^2 y + c$
• D. $\frac{dx}{dy} + x = e^{-y} \sec^2 y$

Hint:

When an equation looks difficult to solve as $\frac{dy}{dx}$, try treating it as $\frac{dx}{dy}$. Linear equations can exist in either form, and swapping the dependent and independent variables often simplifies the $P$ and $Q$ functions.

Answer 1

The Correct Option is D

We need to rearrange the given differential equation into the standard linear form where $x$ is the dependent variable and $y$ is the independent variable ($\frac{dx}{dy} + P(y)x = Q(y)$). Step 1: Divide by $dy$ Starting with: $dx + xdy = e^{-y} \sec^2 y \, dy$ Divide every term by $dy$: $$\frac{dx}{dy} + x = e^{-y} \sec^2 y$$

Step 2: Verification of Linear Form In this form:

• The dependent variable is $x$.

• $P(y) = 1$ (the coefficient of $x$).

• $Q(y) = e^{-y} \sec^2 y$.
This matches the structure of a first-order linear differential equation. Therefore, Option (D) is the correct transformation.