Q.The general solution of a differential equation of the type $\frac{dx}{dy} + P_1 x = Q_1$ is (A) $y e^{\int P_1\, dy} = \int \left(Q_1 e^{\int P_1\, dy}\right) dy + C$ (B) $y e^{\int P_1\, dx} = \int \left(Q_1 e^{\int P_1\, dx}\right) dx + C$ (C) $x e^{\int P_1\, dy} = \int \left(Q_1 e^{\int P_1\, dy}\right) dy + C$ (D) $x e^{\int P_1\, dx} = \int \left(Q_1 e^{\int P_1\, dx}\right) dx + C$

The key idea is that the given equation $\frac{dx}{dy} + P_1 x = Q_1$ is a linear first-order differential equation in $x$ as a function of $y$. The standard integrating factor method applies with the independent variable being $y$.

Step 1: Identify the standard form. For $\frac{dx}{dy} + P_1 x = Q_1$, the integrating factor is $e^{\int P_1\, dy}$.

Step 2: Multiply the entire equation by this integrating factor:

$e^{\int P_1\, dy} \frac{dx}{dy} + P_1 x e^{\int P_1\, dy} = Q_1 e^{\int P_1\, dy}$. …

The key here is to recognise the form of the differential equation. You’re used to seeing $\frac{dy}{dx} + P y = Q$, where $y$ is the dependent variable and $x$ is the independent variable. That’s the standard linear first-order ODE. But the problem flips the roles: here, $x$ depends on $y$, so the derivative is $\frac{dx}{dy}$.

The structure is identical — only the letters have swapped. So the method is exactly the same: find an integrating factor, multiply through, and integrate.

Let’s walk through it.

1. Identify the independent variable.

   The equation is $\frac{dx}{dy} + P_1 x = Q_1$. The derivative is taken with respect to $y$, so $y$ is the independent variable and $x$ is the dependent variable. $P_1$ and $Q_1$ are functions of $y$ (or constants — it doesn’t matter; the method is the same).

2. Recall the standard solution for a linear first-order ODE.

   For $\frac{dy}{dx} + P y = Q$, the integrating factor is $e^{\int P\, dx}$, and the solution is

This is a formula you should know cold. The logic: multiplying the whole equation by the integrating factor turns the left side into the derivative of $(y \cdot \text{IF})$, which you then integrate.

3. Swap variables to match the given form.

   In our problem, $x$ plays the role of $y$, and $y$ plays the role of $x$. So:

   • The dependent variable becomes $x$.
   • The independent variable becomes $y$.
   • The coefficient $P_1$ is a function of $y$.
   • The right-hand side $Q_1$ is also a function of $y$.

   Therefore, the integrating factor becomes $e^{\int P_1\, dy}$, not $e^{\int P_1\, dx}$.

4. Write the general solution by analogy.

   Following the pattern:

which is

$x e^{\int P_1\, dy} = \int \left(Q_1 e^{\int P_1\, dy}\right) dy + C.$ …

Method: The general-solution formula for a $\dfrac{dx}{dy}$ linear equation

For a formula-recognition question, match the integrating-factor solution to the correct independent variable.

Steps

Step 1: Identify the independent variable.

In $\dfrac{dx}{dy}+P_1x=Q_1$, the derivative is with respect to $y$, so $y$ is independent and $x$ dependent.

Step 2: Build the integrating factor in the right variable.

$\text{I.F.}=e^{\int P_1\,dy}\quad(\text{use }dy,\ \text{not }dx).$ …

Common Mistakes

Mistake 1: Using $dx$ instead of $dy$ in the integrating factor.

Why it's wrong: since the derivative is $\dfrac{dx}{dy}$, the independent variable is $y$, so I.F. $=e^{\int P_1\,dy}$; option (D) with $dx$ is the trap. Correct approach: integrate $P_1$ with respect to $y$.

Mistake 2: Putting the wrong variable next to the I.F. …